Simultaneous linear & quadratic equations

1. Revise simultaneous linear equations

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

5x5y=155xy=3
5x5y5xy=15=3

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

HINT: <no title>
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You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
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For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two xx-terms and two yy-terms:

5x5y=155xy=3
5x5y5xy=15=3

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the xx-terms or the yy-terms.

For the equations in this question, notice that the coefficients of xx are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

5x5y=155xy=35x+5x5yy=15+3x-termsdisappear! 0x6y=186y=18
5x5y5xy5x+5x5yyx-termsdisappear! 0x6y6y=15=3=15+3=18=18

The xx-terms cancel each other out - that is what 'elimination' means! We get the 0x0x because the xx-coefficients are equal and opposite: 5x+(5x)5x+(5x) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the xx-terms because the xx-coefficients will cancel.


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Solving simultaneous equations by elimination

Solve for mm and nn:

6m+10n=145m+2n=20
6m+10n5m+2n=14=20
Answer: m=m= and n=n=
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HINT: <no title>
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Look at the coefficients of the nn-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
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The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the nn-term in the first equation is a multiple of the second nn-term: the 22 is 55 times as much as 1010. If we multiply the entire second equation by this factor of 55 we will be able to cancel the nn-terms between the equations:

5m+2n=20(5)5m+(5)2n=20(5)25m+10n=100
5m+2n(5)5m+(5)2n25m+10n=20=20(5)=100
NOTE: We could also multiply the equation by −5−5. That will also make it possible to cancel the nn-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate nn by subtracting the equations
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Now eliminate the nn-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

6m+10n=14Subtract the equationsto cancel the n-terms(25m+10n)=(100)19m+0n=114
Subtract the equationsto cancel the n-terms6m(25m19m+++10n10n)0n===14(100)114

Now we can solve for mm:

m=11419=6
m=11419=6

STEP: Solve for nn
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Finally, use the value of mm to find the value of nn. Remember that we can use either of the equations to do this, we can pick the easier equation. In this case the first equation has fewer negatives than the second equation - in fact it has none. So we will use the first equation.

6m+10n=146(6)+10n=1436+10n=1410n=50n=5
6m+10n6(6)+10n36+10n10nn=14=14=14=50=5

The answers to the pair of equations are m=6m=6 and n=5n=5.


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Simultaneous equations

Solve the following equations simultaneously:

4x+4y=165x4y=14
4x+4y5x4y=16=14
Answer:

The numbers which solve both equations are x=x= and y=y= .

numeric
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HINT: <no title>
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Notice that the yy-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the yy-terms
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We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for xx and for yy, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the yy-coefficients are equal and opposite. If we add the equations together, the yy-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

4x+4y=165x4y=144x5x+4y4y=16+14x+0y=2x=2
4x+4y5x4y4x5x+4y4yx+0yx=16=14=16+14=2=2

STEP: Solve for xx
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Now we have an equation with only one variable, and we can solve it.

x=2x=2
xx=2=2

STEP: Find the value of yy
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So x=2x=2. But we are not done yet: we also need an answer for yy. We can get this using the answer we just got for xx. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the first equation, because it has fewer negative signs.

4x+4y=164(2)+4y=168+4y=164y=24y=6
4x+4y4(2)+4y8+4y4yy=16=16=16=24=6

Now we have the complete answer: the numbers which solve the equations are x=2x=2 and y=6y=6. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=2x=2 and y=6y=6.


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Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

5x+4=2yy+11=2x
5x+4y+11=2y=2x
Answer:

The solution is x=x= and y=y= .

numeric
numeric
HINT: <no title>
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You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
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To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to subtract the 1111 term to the other side of the second equation.

NOTE: We can solve these equations using elimination - it will work. But it will probably be more complicated than substitution.

STEP: Do the substitution and solve for xx
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First rearrange the second equation to set up the substitution:

y+11=2xy=2x11
y+11y=2x=2x11

Now do the substitution and complete the solution to find xx.

5x+4=2y5x+4=2(2x11)5x+4=4x+229x=18x=2
5x+45x+45x+49xx=2y=2(2x11)=4x+22=18=2

Super - we know that x=2x=2.


STEP: Determine the other variable
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Now we can use the xx-value to find the value of yy. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

y+11=2xy+11=2(2)y=7
y+11y+11y=2x=2(2)=7

The answer is the pair of numbers x=2x=2 and y=7y=7. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=2x=2 and y=7y=7.


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Solving simultaneous equations by substitution

Solve for xx and yy using substitution:

5x5y=5and5x10y=3
5x5y=5and5x10y=3

Your answers should be exact (do not round off).

Answer: x=x= and y=y=
numeric
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HINT: <no title>
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You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
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The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of 11 or 11, because that would make our work easier. But that is not the case. So we will just make xx the subject of the first equation:

5x5y=55x=5y+5x=y+1
5x5y5xx=5=5y+5=y+1


STEP: Substitute and solve for yy
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Substitute the expression for xx into the second equation. Then simplify and solve for the value of yy:

5x10y=3substitutein for x5(y+1)10y=35y+5=35y=2y=25
5x10ysubstitutein for x5(y+1)10y5y+55yy=3=3=3=2=25

STEP: Find the value of xx
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Substitute this value of yy back into one of the equations and solve for xx. In this case, the equations are similar: they both have 11 negative sign. We will pick the first equation.

5x5y=55x5(25)=55x=7x=75
5x5y5x5(25)5xx=5=5=7=75

The final answers are x=75x=75 and y=25y=25.


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Equations with fractions

Determine the values of gg and hh which solve these two equations:

2h8=6g3g+16gh=3hg,h0
2h83g+16gh=6g=3hg,h0

Your answer should be exact (do not round off).

Answer:

The solution is g=g= and h=h= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is ghgh.


STEP: Rearrange the second equation
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We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by ghgh, which is the LCD of the fractions. That will cancel all of the denominators.

3g+16gh=3hgh(3g+16gh)=gh(3h)3ggh+16=3hgh3h+16=3g
3g+16ghgh(3g+16gh)3ggh+163h+16=3h=gh(3h)=3hgh=3g

STEP: Pick a method and solve for whichever variable comes more easily
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Now we are ready to solve the following equations:

2h8=6g3h+16=3g
2h8=6g3h+16=3g

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the second equation by 22 so that the gg-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

2(3h+16)=2(3g)6h+32=6g
2(3h+16)6h+32=2(3g)=6g

Now do the elimination and complete the solution to find hh.

2h8=6gSubtract thesecond equation(6h+32)=(6g)328+2h+6h=6g+6g8h40=08h=40h=5
2h8Subtract thesecond equation(6h+32)328+2h+6h8h408hh=6g=(6g)=6g+6g=0=40=5

Great - we have the first value, h=5h=5.


STEP: Solve for the other variable
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Now the last step: find the value of gg. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

2h8=6g2(5)8=6g6g=2g=13
2h82(5)86gg=6g=6g=2=13

The answer is the pair of numbers g=13g=13 and h=5h=5. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force g,h0g,h0, which is why there are open intervals at the gg- and hh-intercepts of the fraction equation.

The values which solve these two equations are g=13g=13 and h=5h=5.


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Solving simultaneous equations by elimination

Solve for xx and yy using the elimination method:

3x+5y=2115x4y=18
3x+5y15x4y=21=18
Answer: x=x= and y=y=
numeric
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HINT: <no title>
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Look at the coefficients of the xx-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the xx-terms
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We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second xx-term has a coefficient which is a multiple of the first xx-term. In fact, 1515 is 55 times as much as 33. We can make both coefficients of xx the same by using this factor of 55 to multiply the entire first equation:

3x+5y=21(5)3x+(5)5y=21(5)15x+25y=105
3x+5y(5)3x+(5)5y15x+25y=21=21(5)=105

STEP: Eliminate the xx-variable and solve for yy
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Now we can eliminate the xx-terms by subtracting one equation from the other. Then we can solve for yy.

15x+25y=105Subtract the equationsto cancel the x-terms:(15x4y)=(18)0x+29y=87
Subtract the equationsto cancel the x-terms:15x(15x0x++25y4y)29y===105(18)87

Now we can solve for yy:

y=8729=3
y=8729=3

STEP: Substitute in for yy
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The last step is to substitute yy back into either of the equations so that we can find xx. Here we will use the first equation:

3x+5y=213x+5(3)=213x15=213x=6x=2
3x+5y3x+5(3)3x153xx=21=21=21=6=2

The answers are x=2x=2 and y=3y=3.


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Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

2x=24yy=3x4
2xy=24y=3x4

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
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For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two xx-terms and two yy-terms:

2x=24yy=3x4
2xy=24y=3x4

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the yy from the second equation into the first equation. This is because in the second equation the yy is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

2x=24yy=3x4
2xy=24y=3x4

Combining these equations by substitution, we get:

2x=24(3x4)
2x=24(3x4)

Since the second equation tells us that yy is equal to y=3x4y=3x4, we can substitute it into the other equation in place of yy straight away.

The best choice is to substitute the yy from the second equation into the first equation because it is already isolated.


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Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

8=4a+4b2=5a3b
82=4a+4b=5a3b

He is using the elimination method. He started by multiplying the second equation by 44, leading to this:

8=4a+4b8=20a+12b
88=4a+4b=20a+12b

But your friend is not sure what to do next, and he asks you for help. What number can you use to multiply the first equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the aa-terms or the bb-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
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For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by 44. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply 44 to get 2020 so we can cancel the aa-terms, or multiply the 44 to get 1212 so we can cancel the bb-terms.

For these equations, both options work. The coefficients of the aa-terms and the bb-terms in the second equation are multiples of the coefficients in the other equation. If we multiply by 55 we can cancel the aa-terms. But we could also multiply by 33 which would allow us to cancel the bb-terms. Both are good choices. Here we will multiply the first equation by 55.

multiply eachterm in the equation8(5)=4(5)a+4(5)b40=20a+20b
multiply eachterm in the equation8(5)40=4(5)a+4(5)b=20a+20b

The coefficients we want to cancel are equal and opposite. So eliminating the terms requires adding both sides of the equation:

40=20a+20badd thisequation8=20a+12b40+8=20a+20a+20b+12b32=0a+32b
40add thisequation840+832=20a+20b=20a+12b=20a+20a+20b+12b=0a+32b
NOTE: If we had multiplied by 55, the signs in the first equation would be changed. Then we would subtract the equations instead. That means that 55 is also an acceptable answer.

The correct answer can be any of these numbers: 55, 55, 33 or 33.


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Picking a solution method

The following equations both include the variables xx and yy:

y=x+42x=14+5y
y2x=x+4=14+5y
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    y=x+42x=14+5y
    y2x=x+4=14+5y

    In this case, the first equation points us to substitution: the yy in that equation is isolated. This is the perfect arrangement for substitution, because we can substitute y=x+4y=x+4 directly into the yy in the second equation.

    The correct answer is: the substitution method.


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  2. Solve the equations simultaneously.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y)(x;y).
    Answer: The answer is: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the method identified in Question 1.


    STEP: Combine the equations using substitution
    [−1 point ⇒ 3 / 4 points left]

    Based on the result of Question 1, we should solve this question using substitution. It is possible to solve the equations using elimination, but using substitution is easier (as described in Question 1). Substitute y=x+4y=x+4 into the second equation in place of yy:

    2x=14+5yy=x+4
    2xy=14+5y=x+4

    Combining these equations by substution, we get:

    2x=14+5(x+4)
    2x=14+5(x+4)

    STEP: Solve for xx
    [−2 points ⇒ 1 / 4 points left]

    Now we can solve the equation. Distribute the 55 and get on with solving the equation.

    2x=14+5(x+4)2x=14+5(x)+5(4)2x=14+5x+202x5x=14+203x=6x=2
    2x2x2x2x5x3xx=14+5(x+4)=14+5(x)+5(4)=14+5x+20=14+20=6=2

    STEP: Find the value of yy
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    So x=2x=2. But remember that we also need an answer for yy. We can find this using the answer we just got for xx. It is important to remember that you can use either equation to calculate yy, because the numbers we want solve both equations. It is good to pick the easiest choice. In this case, the easier choice is the first equation, because it is arranged in a more useful way.

    y=x+4y=(2)+4y=2+4y=2
    yyyy=x+4=(2)+4=2+4=2

    The numbers which solve the equations simultaneously are x=2x=2 and y=2y=2.

    The correct answer is (2;2)(2;2).


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Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=2x1y=x+2
yy=2x1=x+2

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x=x= and y=y= .

numeric
numeric
HINT: <no title>
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You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
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This question is about two equations. We need to find the values of xx and yy which solve the equations 'simultaneously'. This means we want one pair of xx and yy values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (3;5)(3;5).

We can prove that this is correct by substituting the values into each of the equations:

y=2x1(5)=2(3)15=5
y(5)5=2x1=2(3)1=5

and

y=x+2(5)=(3)+25=5
y(5)5=x+2=(3)+2=5

Perfect - the values x=3x=3 and y=5y=5 solve the equations simultaneously.

The correct answers are x=3x=3 and y=5y=5.


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Solving simultaneous equations by substitution

Solve simultaneously for xx and yy.

6x+5y=454x+y=23
6x+5y4x+y=45=23
TIP: Use the substitution method.
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
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When we solve simultaneous equations, we are working to find values for yy and xx which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the yy has a coefficient of one:

4x+y=23target for isolation
4x+y=23target for isolation

STEP: Make yy the subject of the equation
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Isolate yy in the second equation, which can be done in one step:

4x+y=23y=4x23
4x+yy=23=4x23

STEP: Substitute the result into the other equation to find xx
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Now we can substitute the expression 4x234x23 into the first equation and solve:

6x+5y=456x+5(4x23)=456x20x115=4514x=70x=5
6x+5y6x+5(4x23)6x20x11514xx=45=45=45=70=5

STEP: Use the xx-value to find yy
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Right now we have only half of the answer! To get the other half, we need to find the value of yy. We can use the equation we got when we isolated yy: it is the most convenient choice because yy is the subject of the equation.

y=4x23y=4(5)23y=3
yyy=4x23=4(5)23=3

The correct answers are x=5x=5 and y=3y=3.


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Exercises

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

4x+4y=164x+5y=7
4x+4y4x+5y=16=7

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two xx-terms and two yy-terms:

4x+4y=164x+5y=7
4x+4y4x+5y=16=7

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the xx-terms or the yy-terms.

For the equations in this question, notice that the coefficients of xx are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

4x+4y=164x+5y=74x4x+4y+5y=167x-termsdisappear! 0x+9y=99y=9
4x+4y4x+5y4x4x+4y+5yx-termsdisappear! 0x+9y9y=16=7=167=9=9

The xx-terms cancel each other out - that is what 'elimination' means! We get the 0x0x because the xx-coefficients are equal and opposite: 4x+(4x)4x+(4x) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the xx-terms because the xx-coefficients will cancel.


Submit your answer as: and

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

5x4y=205xy=5
5x4y5xy=20=5

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two xx-terms and two yy-terms:

5x4y=205xy=5
5x4y5xy=20=5

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the xx-terms or the yy-terms.

For the equations in this question, notice that the coefficients of xx are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

5x4y=205xy=55x+5x4yy=20+5x-termsdisappear! 0x5y=255y=25
5x4y5xy5x+5x4yyx-termsdisappear! 0x5y5y=20=5=20+5=25=25

The xx-terms cancel each other out - that is what 'elimination' means! We get the 0x0x because the xx-coefficients are equal and opposite: 5x+(5x)5x+(5x) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the xx-terms because the xx-coefficients will cancel.


Submit your answer as: and

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

14=2x+2y17=2x5y
1417=2x+2y=2x5y

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two xx-terms and two yy-terms:

14=2x+2y17=2x5y
1417=2x+2y=2x5y

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the xx-terms or the yy-terms.

For the equations in this question, notice that the coefficients of xx are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

14=2x+2y17=2x5y14+17=2x2x+2y5yx-termsdisappear! 3=0x3y3=3y
141714+17x-termsdisappear! 33=2x+2y=2x5y=2x2x+2y5y=0x3y=3y

The xx-terms cancel each other out - that is what 'elimination' means! We get the 0x0x because the xx-coefficients are equal and opposite: 2x+(2x)2x+(2x) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the xx-terms because the xx-coefficients will cancel.


Submit your answer as: and

Solving simultaneous equations by elimination

Solve for mm and nn:

3m10n=315m+2n=23
3m10n5m+2n=31=23
Answer: m=m= and n=n=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the nn-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
[−2 points ⇒ 4 / 6 points left]

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the nn-term in the first equation is a multiple of the second nn-term: the 22 is 55 times as much as 1010. If we multiply the entire second equation by this factor of 55 we will be able to cancel the nn-terms between the equations:

5m+2n=23(5)5m+(5)2n=23(5)25m+10n=115
5m+2n(5)5m+(5)2n25m+10n=23=23(5)=115
NOTE: We could also multiply the equation by −5−5. That will also make it possible to cancel the nn-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate nn by adding the equations
[−2 points ⇒ 2 / 6 points left]

Now eliminate the nn-terms. Since the terms have opposite signs, we must add one equation to the other to cancel those terms.

3m10n=31Add the equationsto cancel the n-terms+(25m+10n)=+11528m+0n=84
Add the equationsto cancel the n-terms3m+(25m28m++10n10n)0n===31+11584

Now we can solve for mm:

m=8428=3
m=8428=3

STEP: Solve for nn
[−2 points ⇒ 0 / 6 points left]

Finally, use the value of mm to find the value of nn. Remember that we can use either of the equations to do this, we can pick the easier equation. The second equation looks like a better choice because it has fewer negative signs than the first equation (in fact it does not have any!).

5m+2n=235(3)+2n=2315+2n=232n=8n=4
5m+2n5(3)+2n15+2n2nn=23=23=23=8=4

The answers to the pair of equations are m=3m=3 and n=4n=4.


Submit your answer as: and

Solving simultaneous equations by elimination

Solve for jj and kk:

25j+5k=205j+4k=1
25j+5k5j+4k=20=1
Answer: j=j= and k=k=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the jj-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
[−2 points ⇒ 4 / 6 points left]

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the jj-term in the first equation is a multiple of the second jj-term: the 55 is 55 times as much as 2525. If we multiply the entire second equation by this factor of 55 we will be able to cancel the jj-terms between the equations:

5j+4k=1(5)5j+(5)4k=1(5)25j+20k=5
5j+4k(5)5j+(5)4k25j+20k=1=1(5)=5
NOTE: We could also multiply the equation by −5−5. That will also make it possible to cancel the jj-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate jj by subtracting the equations
[−2 points ⇒ 2 / 6 points left]

Now eliminate the jj-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

25j+5k=20Subtract the equationsto cancel the j-terms(25j+20k)=50j15k=15
Subtract the equationsto cancel the j-terms25j(25j0j++5k20k)15k===20515

Now we can solve for kk:

k=1515=1
k=1515=1

STEP: Solve for jj
[−2 points ⇒ 0 / 6 points left]

Finally, use the value of kk to find the value of jj. Remember that we can use either of the equations to do this, we can pick the easier equation. The equations are pretty much the same - for example, neither has any fractions or negatives. So we will use the first one.

25j+5k=2025j+5(1)=2025j5=2025j=25j=1
25j+5k25j+5(1)25j525jj=20=20=20=25=1

The answers to the pair of equations are j=1j=1 and k=1k=1.


Submit your answer as: and

Solving simultaneous equations by elimination

Solve for mm and nn:

5m+16n=92m4n=14
5m+16n2m4n=9=14
Answer: m=m= and n=n=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the nn-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
[−2 points ⇒ 4 / 6 points left]

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the nn-term in the first equation is a multiple of the second nn-term: the 44 is 44 times as much as 1616. If we multiply the entire second equation by this factor of 44 we will be able to cancel the nn-terms between the equations:

2m4n=14(4)2m(4)4n=14(4)8m16n=56
2m4n(4)2m(4)4n8m16n=14=14(4)=56
NOTE: We could also multiply the equation by −4−4. That will also make it possible to cancel the nn-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate nn by adding the equations
[−2 points ⇒ 2 / 6 points left]

Now eliminate the nn-terms. Since the terms have opposite signs, we must add one equation to the other to cancel those terms.

5m+16n=9Add the equationsto cancel the n-terms+(8m16n)=+(56)13m+0n=65
Add the equationsto cancel the n-terms5m+(8m13m++16n16n)0n===9+(56)65

Now we can solve for mm:

m=6513=5
m=6513=5

STEP: Solve for nn
[−2 points ⇒ 0 / 6 points left]

Finally, use the value of mm to find the value of nn. Remember that we can use either of the equations to do this, we can pick the easier equation. In this case the first equation has fewer negatives than the second equation. So we will use the first equation.

5m+16n=95(5)+16n=92516n=916n=16n=1
5m+16n5(5)+16n2516n16nn=9=9=9=16=1

The answers to the pair of equations are m=5m=5 and n=1n=1.


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Simultaneous equations

Solve the following equations simultaneously:

12=5x2y4=3x+2y
124=5x2y=3x+2y
Answer:

The numbers which solve both equations are x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Notice that the yy-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the yy-terms
[−1 point ⇒ 2 / 3 points left]

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for xx and for yy, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the yy-coefficients are equal and opposite. If we add the equations together, the yy-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

12=5x2y4=3x+2y12+4=5x3x2y+2y16=8x+0y16=8x
12412+41616=5x2y=3x+2y=5x3x2y+2y=8x+0y=8x

STEP: Solve for xx
[−1 point ⇒ 1 / 3 points left]

Now we have an equation with only one variable, and we can solve it.

16=8x2=x
162=8x=x

STEP: Find the value of yy
[−1 point ⇒ 0 / 3 points left]

So x=2x=2. But we are not done yet: we also need an answer for yy. We can get this using the answer we just got for xx. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the second equation, because it has fewer negative signs.

4=3x+2y4=3(2)+2y4=6+2y2=2y1=y
44421=3x+2y=3(2)+2y=6+2y=2y=y

Now we have the complete answer: the numbers which solve the equations are x=2x=2 and y=1y=1. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=2x=2 and y=1y=1.


Submit your answer as: and

Simultaneous equations

Solve the following equations simultaneously:

4=4x4y14=4xy
414=4x4y=4xy
Answer:

The numbers which solve both equations are x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Notice that the xx-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the xx-terms
[−1 point ⇒ 2 / 3 points left]

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for xx and for yy, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the xx-coefficients are equal and opposite. If we add the equations together, the xx-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

4=4x4y14=4xy4+14=4x4x4yy10=0x5y10=5y
4144+141010=4x4y=4xy=4x4x4yy=0x5y=5y

STEP: Solve for yy
[−1 point ⇒ 1 / 3 points left]

Now we have an equation with only one variable, and we can solve it.

10=5y2=y
102=5y=y

STEP: Find the value of xx
[−1 point ⇒ 0 / 3 points left]

So y=2y=2. But we are not done yet: we also need an answer for xx. We can get this using the answer we just got for yy. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. In this case, neither of the equations is much more friendly than the other, so we will just use the first equation.

4=4x4y4=4x4(2)4=4x+812=4x3=x
444123=4x4y=4x4(2)=4x+8=4x=x

Now we have the complete answer: the numbers which solve the equations are x=3x=3 and y=2y=2. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=3x=3 and y=2y=2.


Submit your answer as: and

Simultaneous equations

Solve the following equations simultaneously:

3x+4y=18x4y=6
3x+4yx4y=18=6
Answer:

The numbers which solve both equations are x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Notice that the yy-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the yy-terms
[−1 point ⇒ 2 / 3 points left]

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for xx and for yy, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the yy-coefficients are equal and opposite. If we add the equations together, the yy-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

3x+4y=18x4y=63x+x+4y4y=1864x+0y=244x=24
3x+4yx4y3x+x+4y4y4x+0y4x=18=6=186=24=24

STEP: Solve for xx
[−1 point ⇒ 1 / 3 points left]

Now we have an equation with only one variable, and we can solve it.

4x=24x=6
4xx=24=6

STEP: Find the value of yy
[−1 point ⇒ 0 / 3 points left]

So x=6x=6. But we are not done yet: we also need an answer for yy. We can get this using the answer we just got for xx. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the first equation, because it has fewer negative signs.

3x+4y=183(6)+4y=1818+4y=184y=0y=0
3x+4y3(6)+4y18+4y4yy=18=18=18=0=0

Now we have the complete answer: the numbers which solve the equations are x=6x=6 and y=0y=0. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=6x=6 and y=0y=0.


Submit your answer as: and

Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

5x=y+103x4y=6
5x3x4y=y+10=6
Answer:

The solution is x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
[−1 point ⇒ 5 / 6 points left]

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to subtract the 1010 term to the other side of the first equation.

NOTE: We can solve these equations using elimination - it will work. But it will probably be more complicated than substitution.

STEP: Do the substitution and solve for xx
[−4 points ⇒ 1 / 6 points left]

First rearrange the first equation to set up the substitution:

5x=y+105x10=y
5x5x10=y+10=y

Now do the substitution and complete the solution to find xx.

3x4y=63x4(5x10)=617x+40=617x=34x=2
3x4y3x4(5x10)17x+4017xx=6=6=6=34=2

Super - we know that x=2x=2.


STEP: Determine the other variable
[−1 point ⇒ 0 / 6 points left]

Now we can use the xx-value to find the value of yy. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

3x4y=63(2)4y=64y=0y=0
3x4y3(2)4y4yy=6=6=0=0

The answer is the pair of numbers x=2x=2 and y=0y=0. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=2x=2 and y=0y=0.


Submit your answer as: and

Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

11=5x+4y6y+12=2x
116y+12=5x+4y=2x
Answer:

The solution is x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
[−1 point ⇒ 5 / 6 points left]

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations are not arranged nicely for elimination or substitution. However, notice that the coefficients in the second equation are all a multiple of 22. So we can divide the equation by 22 to isolate xx without picking up any fractions!

NOTE: We can solve these equations using elimination - it will work. But it will probably be more complicated than substitution.

STEP: Do the substitution and solve for yy
[−4 points ⇒ 1 / 6 points left]

First rearrange the second equation to set up the substitution:

6y+12=2x3y+6=x
6y+123y+6=2x=x

Now do the substitution and complete the solution to find yy.

11=5x+4y11=5(3y+6)+4y11=19y3019y=19y=1
11111119yy=5x+4y=5(3y+6)+4y=19y30=19=1

Super - we know that y=1y=1.


STEP: Determine the other variable
[−1 point ⇒ 0 / 6 points left]

Now we can use the yy-value to find the value of xx. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

6y+12=2x6(1)+12=2x2x=6x=3
6y+126(1)+122xx=2x=2x=6=3

The answer is the pair of numbers x=3x=3 and y=1y=1. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=3x=3 and y=1y=1.


Submit your answer as: and

Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

5y=3x+25=3x2y
5y5=3x+2=3x2y
Answer:

The solution is x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
[−1 point ⇒ 4 / 5 points left]

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations invite elimination, because the xx-terms have equal coefficients. We can eliminate the xx-terms by subtracting the equations.

NOTE: We can solve these equations using substitution - it will work. But it will probably be more complicated than elimination.

STEP: Do the elimination and solve for yy
[−3 points ⇒ 1 / 5 points left]

Eliminate the xx and solve for yy:

5y=3x+2Subtract thesecond equation(5)=(3x2y)55y=23x+3x+2y5y5=2y+27y=7y=1
5ySubtract thesecond equation(5)55y5y57yy=3x+2=(3x2y)=23x+3x+2y=2y+2=7=1

Super - we know that y=1y=1.


STEP: Determine the other variable
[−1 point ⇒ 0 / 5 points left]

Now we can use the yy-value to find the value of xx. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case the equations are pretty similar, so we will just pick the first one.

5y=3x+25(1)=3x+23x=3x=1
5y5(1)3xx=3x+2=3x+2=3=1

The answer is the pair of numbers x=1x=1 and y=1y=1. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=1x=1 and y=1y=1.


Submit your answer as: and

Solving simultaneous equations by substitution

Solve for xx and yy using substitution:

4x+6y=10and10x9y=5
4x+6y=10and10x9y=5

Your answers should be exact (do not round off).

Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
[−1 point ⇒ 3 / 4 points left]

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of 11 or 11, because that would make our work easier. But that is not the case. So we will just make xx the subject of the first equation:

4x+6y=104x=6y+10x=3y2+52
4x+6y4xx=10=6y+10=3y2+52

Unfortunately, this equation has 22 fractions in it. That is ugly, but it will work.


STEP: Substitute and solve for yy
[−2 points ⇒ 1 / 4 points left]

Substitute the expression for xx into the second equation. Then simplify and solve for the value of yy:

10x9y=5substitutein for x10(3y2+52)9y=524y+25=524y=20y=56
10x9ysubstitutein for x10(3y2+52)9y24y+2524yy=5=5=5=20=56

STEP: Find the value of xx
[−1 point ⇒ 0 / 4 points left]

Substitute this value of yy back into one of the equations and solve for xx. In this case, we should use the first equation because it has no negative signs.

4x+6y=104x+6(56)=104x=5x=54
4x+6y4x+6(56)4xx=10=10=5=54

The final answers are x=54x=54 and y=56y=56.


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Solving simultaneous equations by substitution

Solve for xx and yy using substitution:

3x4y=7andx8y=5
3x4y=7andx8y=5

Your answers should be exact (do not round off).

Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
[−1 point ⇒ 3 / 4 points left]

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. Luckily for us, we can isolate xx in the second equation pretty easily because the coefficient of that term is 11.

x8y=5x=8y5
x8yx=5=8y5


STEP: Substitute and solve for yy
[−2 points ⇒ 1 / 4 points left]

Substitute the expression for xx into the first equation. Then simplify and solve for the value of yy:

3x4y=7substitutein for x3(8y5)4y=728y+15=728y=8y=27
3x4ysubstitutein for x3(8y5)4y28y+1528yy=7=7=7=8=27

STEP: Find the value of xx
[−1 point ⇒ 0 / 4 points left]

Substitute this value of yy back into one of the equations and solve for xx. In this case, the equations are similar: they both have 22 negative signs. We will pick the first equation.

3x4y=73x4(27)=73x=577x=197
3x4y3x4(27)3xx=7=7=577=197

The final answers are x=197x=197 and y=27y=27.


Submit your answer as: and

Solving simultaneous equations by substitution

Solve for xx and yy using substitution:

x+5y=1and2x3y=8
x+5y=1and2x3y=8

Your answers should be exact (do not round off).

Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
[−1 point ⇒ 3 / 4 points left]

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. Luckily for us, we can isolate xx in the first equation pretty easily because the coefficient of that term is 11.

x+5y=1x=5y+1
x+5yx=1=5y+1


STEP: Substitute and solve for yy
[−2 points ⇒ 1 / 4 points left]

Substitute the expression for xx into the second equation. Then simplify and solve for the value of yy:

2x3y=8substitutein for x2(5y+1)3y=813y+2=813y=10y=1013
2x3ysubstitutein for x2(5y+1)3y13y+213yy=8=8=8=10=1013

STEP: Find the value of xx
[−1 point ⇒ 0 / 4 points left]

Substitute this value of yy back into one of the equations and solve for xx. In this case, we should use the first equation because it has no negative signs.

x+5y=1x+5(1013)=1x=3713
x+5yx+5(1013)x=1=1=3713

The final answers are x=3713x=3713 and y=1013y=1013.


Submit your answer as: and

Equations with fractions

Solve the following equations simultaneously to find the values of gg and hh:

6gh=2h1gg,h05g+20=2h
6gh5g+20=2h1gg,h0=2h

Your answer should be exact (do not round off).

Answer:

The solution is g=g= and h=h= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is ghgh.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by ghgh, which is the LCD of the fractions. That will cancel all of the denominators.

6gh=2h1ggh(6gh)=gh(2h1g)6=1ggh+2hgh6=2gh
6ghgh(6gh)66=2h1g=gh(2h1g)=1ggh+2hgh=2gh

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

6=2gh5g+20=2h
6=2gh5g+20=2h

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 22 so that the hh-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

2(6)=2(2gh)12=4g2h
2(6)12=2(2gh)=4g2h

Now do the elimination and complete the solution to find gg.

12=4g2hSubtract thesecond equation(5g+20)=(2h)20+12+5g=4g2h+2h5g8=4gg=8
12Subtract thesecond equation(5g+20)20+12+5g5g8g=4g2h=(2h)=4g2h+2h=4g=8

Great - we have the first value, g=8g=8.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of hh. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

5g+20=2h5(8)+20=2h2h=20h=10
5g+205(8)+202hh=2h=2h=20=10

The answer is the pair of numbers g=8g=8 and h=10h=10. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force g,h0g,h0, which is why there are open intervals at the gg- and hh-intercepts of the fraction equation.

The values which solve these two equations are g=8g=8 and h=10h=10.


Submit your answer as: and

Equations with fractions

Determine the values of aa and bb which solve these two equations:

5a7ab=1ba,b07=4a+6b
5a7ab7=1ba,b0=4a+6b

Your answer should be exact (do not round off).

Answer:

The solution is a=a= and b=b= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is abab.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by abab, which is the LCD of the fractions. That will cancel all of the denominators.

5a7ab=1bab(5a7ab)=ab(1b)5aab7=1bab5b7=a
5a7abab(5a7ab)5aab75b7=1b=ab(1b)=1bab=a

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

5b7=a7=4a+6b
5b7=a7=4a+6b

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to multiply both sides of the equation by 11 to remove the negative sign from aa.

First rearrange things to set up the substitution:

5b7=a5b+7=a
5b75b+7=a=a

Now do the substitution and complete the solution to find bb.

7=4a+6b7=4(5b+7)+6b7=14b2814b=21b=32
77714bb=4a+6b=4(5b+7)+6b=14b28=21=32

Great - we have the first value, b=32b=32.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of aa. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

7=4a+6b7=4a+6(32)4a=2a=12
774aa=4a+6b=4a+6(32)=2=12

The answer is the pair of numbers a=12a=12 and b=32b=32. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force a,b0a,b0, which is why there are open intervals at the aa- and bb-intercepts of the fraction equation.

The values which solve these two equations are a=12a=12 and b=32b=32.


Submit your answer as: and

Equations with fractions

What are the values of aa and bb which solve these equations simultaneously?

4a6ab=1ba,b03b+8=2a
4a6ab3b+8=1ba,b0=2a

Your answer should be exact (do not round off).

Answer:

The solution is a=a= and b=b= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is abab.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by abab, which is the LCD of the fractions. That will cancel all of the denominators.

4a6ab=1bab(4a6ab)=ab(1b)4aab6=1bab4b6=a
4a6abab(4a6ab)4aab64b6=1b=ab(1b)=1bab=a

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

4b6=a3b+8=2a
4b6=a3b+8=2a

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to multiply both sides of the equation by 11 to remove the negative sign from aa.

First rearrange things to set up the substitution:

4b6=a4b+6=a
4b64b+6=a=a

Now do the substitution and complete the solution to find bb.

3b+8=2a3b+8=2(4b+6)3b+8=8b125b=20b=4
3b+83b+83b+85bb=2a=2(4b+6)=8b12=20=4

Great - we have the first value, b=4b=4.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of aa. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

3b+8=2a3(4)+8=2a2a=20a=10
3b+83(4)+82aa=2a=2a=20=10

The answer is the pair of numbers a=10a=10 and b=4b=4. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force a,b0a,b0, which is why there are open intervals at the aa- and bb-intercepts of the fraction equation.

The values which solve these two equations are a=10a=10 and b=4b=4.


Submit your answer as: and

Solving simultaneous equations by elimination

Solve for xx and yy using the elimination method:

5x+3y=272x+6y=30
5x+3y2x+6y=27=30
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the yy-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the yy-terms
[−2 points ⇒ 4 / 6 points left]

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second yy-term has a coefficient which is a multiple of the first yy-term. In fact, 66 is 22 times as much as 33. We can make both coefficients of yy the same by using this factor of 22 to multiply the entire first equation:

5x+3y=27(2)(5x)+(2)3y=27(2)10x+6y=54
5x+3y(2)(5x)+(2)3y10x+6y=27=27(2)=54

STEP: Eliminate the yy-variable and solve for xx
[−2 points ⇒ 2 / 6 points left]

Now we can eliminate the yy-terms by subtracting one equation from the other. Then we can solve for xx.

10x+6y=54Subtract the equationsto cancel the y-terms:(2x+6y)=(30)8x+0y=24
Subtract the equationsto cancel the y-terms:10x(2x8x+++6y6y)0y===54(30)24

Now we can solve for xx:

x=248=3
x=248=3

STEP: Substitute in for xx
[−2 points ⇒ 0 / 6 points left]

The last step is to substitute yy back into either of the equations so that we can find xx. Here we will use the first equation:

5x+3y=275(3)+3y=2715+3y=273y=12y=4
5x+3y5(3)+3y15+3y3yy=27=27=27=12=4

The answers are x=3x=3 and y=4y=4.


Submit your answer as: and

Solving simultaneous equations by elimination

Solve for xx and yy using the elimination method:

2x3y=64x5y=14
2x3y4x5y=6=14
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the xx-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the xx-terms
[−2 points ⇒ 4 / 6 points left]

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second xx-term has a coefficient which is a multiple of the first xx-term. In fact, 44 is 22 times as much as 22. We can make both coefficients of xx the same by using this factor of 22 to multiply the entire first equation:

2x3y=6(2)2x(2)3y=6(2)4x6y=12
2x3y(2)2x(2)3y4x6y=6=6(2)=12

STEP: Eliminate the xx-variable and solve for yy
[−2 points ⇒ 2 / 6 points left]

Now we can eliminate the xx-terms by subtracting one equation from the other. Then we can solve for yy.

4x6y=12Subtract the equationsto cancel the x-terms:(4x5y)=(14)0xy=2
Subtract the equationsto cancel the x-terms:4x(4x0x6y5y)y===12(14)2

Now we can solve for yy:

y=21=2
y=21=2

STEP: Substitute in for yy
[−2 points ⇒ 0 / 6 points left]

The last step is to substitute yy back into either of the equations so that we can find xx. Here we will use the first equation:

2x3y=62x3(2)=62x6=62x=12x=6
2x3y2x3(2)2x62xx=6=6=6=12=6

The answers are x=6x=6 and y=2y=2.


Submit your answer as: and

Solving simultaneous equations by elimination

Solve for xx and yy using the elimination method:

x+5y=510x4y=4
x+5y10x4y=5=4
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the xx-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the xx-terms
[−2 points ⇒ 4 / 6 points left]

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second xx-term has a coefficient which is a multiple of the first xx-term. In fact, 1010 is 1010 times as much as 11. We can make both coefficients of xx the same by using this factor of 1010 to multiply the entire first equation:

x+5y=5(10)x+(10)5y=5(10)10x+50y=50
x+5y(10)x+(10)5y10x+50y=5=5(10)=50

STEP: Eliminate the xx-variable and solve for yy
[−2 points ⇒ 2 / 6 points left]

Now we can eliminate the xx-terms by subtracting one equation from the other. Then we can solve for yy.

10x+50y=50Subtract the equationsto cancel the x-terms:(10x4y)=(4)0x+54y=54
Subtract the equationsto cancel the x-terms:10x(10x0x++50y4y)54y===50(4)54

Now we can solve for yy:

y=5454=1
y=5454=1

STEP: Substitute in for yy
[−2 points ⇒ 0 / 6 points left]

The last step is to substitute yy back into either of the equations so that we can find xx. Here we will use the first equation:

x+5y=5x+5(1)=5x5=5x=0
x+5yx+5(1)x5x=5=5=5=0

The answers are x=0x=0 and y=1y=1.


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Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

4x=93yy=5x+3
4xy=93y=5x+3

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two xx-terms and two yy-terms:

4x=93yy=5x+3
4xy=93y=5x+3

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the yy from the second equation into the first equation. This is because in the second equation the yy is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

4x=93yy=5x+3
4xy=93y=5x+3

Combining these equations by substitution, we get:

4x=93(5x+3)
4x=93(5x+3)

Since the second equation tells us that yy is equal to y=5x+3y=5x+3, we can substitute it into the other equation in place of yy straight away.

The best choice is to substitute the yy from the second equation into the first equation because it is already isolated.


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Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

4y=204xy=3x1
4yy=204x=3x1

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two xx-terms and two yy-terms:

4y=204xy=3x1
4yy=204x=3x1

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the yy from the second equation into the first equation. This is because in the second equation the yy is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

4y=204xy=3x1
4yy=204x=3x1

Combining these equations by substitution, we get:

4(3x1)=204x
4(3x1)=204x

Since the second equation tells us that yy is equal to y=3x1y=3x1, we can substitute it into the other equation in place of yy straight away.

The best choice is to substitute the yy from the second equation into the first equation because it is already isolated.


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Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

3y=0+3xx=4y5
3yx=0+3x=4y5

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two xx-terms and two yy-terms:

3y=0+3xx=4y5
3yx=0+3x=4y5

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the xx from the second equation into the first equation. This is because in the second equation the xx is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

3y=0+3xx=4y5
3yx=0+3x=4y5

Combining these equations by substitution, we get:

3y=0+3(4y5)
3y=0+3(4y5)

Since the second equation tells us that xx is equal to x=4y5x=4y5, we can substitute it into the other equation in place of xx straight away.

The best choice is to substitute the xx from the second equation into the first equation because it is already isolated.


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Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

3k+5p=92k+3p=6
3k+5p2k+3p=9=6

He is using the elimination method. He started by multiplying the first equation by 33, leading to this:

9k+15p=272k+3p=6
9k+15p2k+3p=27=6

But your friend is not sure what to do next, and he asks you for help. What number can you use to multiply the second equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the kk-terms or the pp-terms. So you need to change one of the coefficients in the second equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
[−1 point ⇒ 0 / 1 points left]

For this question we need to multiply the second equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The first equation was already multiplied by 33. We need to change one of the coefficients in the second equation so that we can cancel terms.

Specifically, we need to multiply 22 to get 99 so we can cancel the kk-terms, or multiply the 33 to get 1515 so we can cancel the pp-terms.

The numbers in these equations point us toward cancelling the pp-terms. This is because 1515 is a multiple of 33. So if we multiply the second equation by 55 the the coefficients will be equal.

multiply eachterm in the equation2(5)k+3(5)p=6(5)10k+15p=30
multiply eachterm in the equation2(5)k+3(5)p10k+15p=6(5)=30

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

9k+15p=27subtract thisequation10k+15p=309k10k+15p15p=27+3019k+0p=57
9k+15psubtract thisequation10k+15p9k10k+15p15p19k+0p=27=30=27+30=57
NOTE: If we had multiplied by 55, the signs in the second equation would be changed. Then we would add the equations instead. That means that 55 is also an acceptable answer.

The correct answer can be either of these numbers: 55 or 55.


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Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

5m+4n=52m+3n=9
5m+4n2m+3n=5=9

She is using the elimination method. She started by multiplying the second equation by 55, leading to this:

5m+4n=510m+15n=45
5m+4n10m+15n=5=45

But your friend is not sure what to do next, and she asks you for help. What number can you use to multiply the first equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the mm-terms or the nn-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
[−1 point ⇒ 0 / 1 points left]

For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by 55. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply 55 to get 1010 so we can cancel the mm-terms, or multiply the 44 to get 1515 so we can cancel the nn-terms.

The numbers in these equations point us toward cancelling the mm-terms. This is because 1010 is a multiple of 55. So if we multiply the first equation by 22 the the coefficients will be equal.

multiply eachterm in the equation5(2)m+4(2)n=5(2)10m+8n=10
multiply eachterm in the equation5(2)m+4(2)n10m+8n=5(2)=10

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

10m+8n=10subtract thisequation10m+15n=4510m10m+8n15n=10450m7n=35
10m+8nsubtract thisequation10m+15n10m10m+8n15n0m7n=10=45=1045=35
NOTE: If we had multiplied by 22, the signs in the first equation would be changed. Then we would add the equations instead. That means that 22 is also an acceptable answer.

The correct answer can be either of these numbers: 22 or 22.


Submit your answer as:

Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

3x+5y=55x+2y=2
3x+5y5x+2y=5=2

He is using the elimination method. He started by multiplying the first equation by 55, leading to this:

15x+25y=255x+2y=2
15x+25y5x+2y=25=2

But your friend is not sure what to do next, and he asks you for help. What number can you use to multiply the second equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the xx-terms or the yy-terms. So you need to change one of the coefficients in the second equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
[−1 point ⇒ 0 / 1 points left]

For this question we need to multiply the second equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The first equation was already multiplied by 55. We need to change one of the coefficients in the second equation so that we can cancel terms.

Specifically, we need to multiply 55 to get 1515 so we can cancel the xx-terms, or multiply the 22 to get 2525 so we can cancel the yy-terms.

The numbers in these equations point us toward cancelling the xx-terms. This is because 1515 is a multiple of 55. So if we multiply the second equation by 33 the the coefficients will be equal and opposite.

multiply eachterm in the equation5(3)x+2(3)y=2(3)15x+6y=6
multiply eachterm in the equation5(3)x+2(3)y15x+6y=2(3)=6

The coefficients we want to cancel are equal and opposite. So eliminating the terms requires adding both sides of the equation:

15x+25y=25add thisequation15x+6y=615x+15x+25y+6y=25+60x+3y=31
15x+25yadd thisequation15x+6y15x+15x+25y+6y0x+3y=25=6=25+6=31
NOTE: If we had multiplied by 33, the signs in the second equation would be changed. Then we would subtract the equations instead. That means that 33 is also an acceptable answer.

The correct answer can be either of these numbers: 33 or 33.


Submit your answer as:

Picking a solution method

These two equations contain the same variables:

y=4x53y=5+2x
y3y=4x5=5+2x
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    y=4x53y=5+2x
    y3y=4x5=5+2x

    In this case, the first equation points us to substitution: the yy in that equation is isolated. This is the perfect arrangement for substitution, because we can substitute y=4x5y=4x5 directly into the yy in the second equation.

    The correct answer is: the substitution method.


    Submit your answer as:
  2. Now find the simultaneous solution to the equations.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y)(x;y).
    Answer: The answer is: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the method identified in Question 1.


    STEP: Combine the equations using substitution
    [−1 point ⇒ 3 / 4 points left]

    Based on the result of Question 1, we should solve this question using substitution. It is possible to solve the equations using elimination, but using substitution is easier (as described in Question 1). Substitute y=4x5y=4x5 into the second equation in place of yy:

    3y=5+2xy=4x5
    3yy=5+2x=4x5

    Combining these equations by substution, we get:

    3(4x5)=5+2x
    3(4x5)=5+2x

    STEP: Solve for xx
    [−2 points ⇒ 1 / 4 points left]

    Now we can solve the equation. Distribute the 33 and get on with solving the equation.

    3(4x5)=5+2x3(4x)3(5)=5+2x12x+15=5+2x2x+12x=51510x=10x=1
    3(4x5)3(4x)3(5)12x+152x+12x10xx=5+2x=5+2x=5+2x=515=10=1

    STEP: Find the value of yy
    [−1 point ⇒ 0 / 4 points left]

    So x=1x=1. But remember that we also need an answer for yy. We can find this using the answer we just got for xx. It is important to remember that you can use either equation to calculate yy, because the numbers we want solve both equations. It is good to pick the easiest choice. In this case, the easier choice is the first equation, because it is arranged in a more useful way.

    y=4x5y=4(1)5y=45y=1
    yyyy=4x5=4(1)5=45=1

    The numbers which solve the equations simultaneously are x=1x=1 and y=1y=1.

    The correct answer is (1;1)(1;1).


    Submit your answer as:

Picking a solution method

Here are two equations, which contain the same variables:

3x=17+4yy=x+5
3xy=17+4y=x+5
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    3x=17+4yy=x+5
    3xy=17+4y=x+5

    In this case, the second equation points us to substitution: the yy in that equation is isolated. This is the perfect arrangement for substitution, because we can substitute y=x+5y=x+5 directly into the yy in the first equation.

    The correct answer is: the substitution method.


    Submit your answer as:
  2. Now solve the equations simultaneously.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y)(x;y).
    Answer: The answer is: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the method identified in Question 1.


    STEP: Combine the equations using substitution
    [−1 point ⇒ 3 / 4 points left]

    Based on the result of Question 1, we should solve this question using substitution. It is possible to solve the equations using elimination, but using substitution is easier (as described in Question 1). Substitute y=x+5y=x+5 into the first equation in place of yy:

    3x=17+4yy=x+5
    3xy=17+4y=x+5

    Combining these equations by substution, we get:

    3x=17+4(x+5)
    3x=17+4(x+5)

    STEP: Solve for xx
    [−2 points ⇒ 1 / 4 points left]

    Now we can solve the equation. Distribute the 44 and get on with solving the equation.

    3x=17+4(x+5)3x=17+4(x)+4(5)3x=174x+203x+4x=17+20x=3
    3x3x3x3x+4xx=17+4(x+5)=17+4(x)+4(5)=174x+20=17+20=3

    STEP: Find the value of yy
    [−1 point ⇒ 0 / 4 points left]

    So x=3x=3. But remember that we also need an answer for yy. We can find this using the answer we just got for xx. It is important to remember that you can use either equation to calculate yy, because the numbers we want solve both equations. It is good to pick the easiest choice. In this case, the easier choice is the second equation, because it is arranged in a more useful way.

    y=x+5y=(3)+5y=3+5y=2
    yyyy=x+5=(3)+5=3+5=2

    The numbers which solve the equations simultaneously are x=3x=3 and y=2y=2.

    The correct answer is (3;2)(3;2).


    Submit your answer as:

Picking a solution method

The following equations both include the variables xx and yy:

3x+3y=64x3y=15
3x+3y4x3y=6=15
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    3x+3y=64x3y=15
    3x+3y4x3y=6=15

    In this case, the yy-terms are ready to be eliminated. The coefficients of these terms are equal and opposite, which is perfect for the elimination method: it means we can cancel those two terms if we add the equations, which is how elimination works.

    The correct answer is: the elimination method.


    Submit your answer as:
  2. Now solve the equations simultaneously.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y)(x;y).
    Answer: The answer is: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the method identified in Question 1.


    STEP: Eliminate (cancel) the yy-terms
    [−1 point ⇒ 2 / 3 points left]

    Based on the result of Question 1, we should solve this question using elimination. It is possible to solve the equations using substitution, but using elimination is easier (as described in Question 1). Add the equations together to eliminate the yy-terms. This means we add together the left and right sides of the equations.

    3x+3y=6add thisequation4x3y=153x+4x+3y3y=6+157x+0y=217x=21
    3x+3yadd thisequation4x3y3x+4x+3y3y7x+0y7x=6=15=6+15=21=21


    STEP: Solve for xx
    [−1 point ⇒ 1 / 3 points left]

    Now we have an equation with only one variable, and we can solve it.

    7x=21x=3
    7xx=21=3

    STEP: Find the value of yy
    [−1 point ⇒ 0 / 3 points left]

    So x=3x=3. But we also need an answer for yy. We can get this using the answer we just got for xx: substitute this into either of the equations to get the answer. We can use either equation because we are looking for the number that solves both of them. Since we have a choice, we should pick the easier equation (if there is one). We will pick the first equation, because it has no negative signs.

    3x+3y=63(3)+3y=69+3y=63y=3y=1
    3x+3y3(3)+3y9+3y3yy=6=6=6=3=1

    Now we have the complete answer: the numbers which solve the equations simultaneously are x=3x=3 and y=1y=1.

    The correct answer is (3;1)(3;1).


    Submit your answer as:

Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=3xy=2x+5
yy=3x=2x+5

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
[−2 points ⇒ 0 / 2 points left]

This question is about two equations. We need to find the values of xx and yy which solve the equations 'simultaneously'. This means we want one pair of xx and yy values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (1;3)(1;3).

We can prove that this is correct by substituting the values into each of the equations:

y=3x(3)=3(1)3=3
y(3)3=3x=3(1)=3

and

y=2x+5(3)=2(1)+53=3
y(3)3=2x+5=2(1)+5=3

Perfect - the values x=1x=1 and y=3y=3 solve the equations simultaneously.

The correct answers are x=1x=1 and y=3y=3.


Submit your answer as: and

Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=2x3+4y=x3+1
yy=2x3+4=x3+1

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
[−2 points ⇒ 0 / 2 points left]

This question is about two equations. We need to find the values of xx and yy which solve the equations 'simultaneously'. This means we want one pair of xx and yy values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (3;2)(3;2).

We can prove that this is correct by substituting the values into each of the equations:

y=2x3+4(2)=2(3)3+42=2
y(2)2=2x3+4=2(3)3+4=2

and

y=x3+1(2)=(3)3+12=2
y(2)2=x3+1=(3)3+1=2

Perfect - the values x=3x=3 and y=2y=2 solve the equations simultaneously.

The correct answers are x=3x=3 and y=2y=2.


Submit your answer as: and

Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=x+1y=3x5
yy=x+1=3x5

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x=x= and y=y= .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
[−2 points ⇒ 0 / 2 points left]

This question is about two equations. We need to find the values of xx and yy which solve the equations 'simultaneously'. This means we want one pair of xx and yy values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (3;4)(3;4).

We can prove that this is correct by substituting the values into each of the equations:

y=x+1(4)=(3)+14=4
y(4)4=x+1=(3)+1=4

and

y=3x5(4)=3(3)54=4
y(4)4=3x5=3(3)5=4

Perfect - the values x=3x=3 and y=4y=4 solve the equations simultaneously.

The correct answers are x=3x=3 and y=4y=4.


Submit your answer as: and

Solving simultaneous equations by substitution

Solve simultaneously for xx and yy.

3x2y=28x+5y=19
3x2yx+5y=28=19
TIP: Use the substitution method.
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
[−1 point ⇒ 5 / 6 points left]

When we solve simultaneous equations, we are working to find values for xx and yy which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the xx has a coefficient of one:

x+5y=19target for isolation
x+5y=19target for isolation

STEP: Make xx the subject of the equation
[−1 point ⇒ 4 / 6 points left]

Isolate xx in the second equation, which can be done in one step:

x+5y=19x=5y19
x+5yx=19=5y19

STEP: Substitute the result into the other equation to find yy
[−2 points ⇒ 2 / 6 points left]

Now we can substitute the expression 5y195y19 into the first equation and solve:

3x2y=283(5y19)2y=2815y572y=2817y=85y=5
3x2y3(5y19)2y15y572y17yy=28=28=28=85=5

STEP: Use the yy-value to find xx
[−2 points ⇒ 0 / 6 points left]

Right now we have only half of the answer! To get the other half, we need to find the value of xx. We can use the equation we got when we isolated xx: it is the most convenient choice because xx is the subject of the equation.

x=5y19x=5(5)19x=6
xxx=5y19=5(5)19=6

The correct answers are x=6x=6 and y=5y=5.


Submit your answer as: and

Solving simultaneous equations by substitution

Solve simultaneously for xx and yy.

4x3y=48x+4y=7
4x3yx+4y=48=7
TIP: Use the substitution method.
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
[−1 point ⇒ 5 / 6 points left]

When we solve simultaneous equations, we are working to find values for xx and yy which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the xx has a coefficient of one:

x+4y=7target for isolation
x+4y=7target for isolation

STEP: Make xx the subject of the equation
[−1 point ⇒ 4 / 6 points left]

Isolate xx in the second equation, which can be done in one step:

x+4y=7x=4y7
x+4yx=7=4y7

STEP: Substitute the result into the other equation to find yy
[−2 points ⇒ 2 / 6 points left]

Now we can substitute the expression 4y74y7 into the first equation and solve:

4x3y=484(4y7)3y=4816y283y=4819y=76y=4
4x3y4(4y7)3y16y283y19yy=48=48=48=76=4

STEP: Use the yy-value to find xx
[−2 points ⇒ 0 / 6 points left]

Right now we have only half of the answer! To get the other half, we need to find the value of xx. We can use the equation we got when we isolated xx: it is the most convenient choice because xx is the subject of the equation.

x=4y7x=4(4)7x=9
xxx=4y7=4(4)7=9

The correct answers are x=9x=9 and y=4y=4.


Submit your answer as: and

Solving simultaneous equations by substitution

Solve simultaneously for xx and yy.

3x+4y=2x6y=36
3x+4yx6y=2=36
TIP: Use the substitution method.
Answer: x=x= and y=y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
[−1 point ⇒ 5 / 6 points left]

When we solve simultaneous equations, we are working to find values for xx and yy which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the xx has a coefficient of one:

x6y=36target for isolation
x6y=36target for isolation

STEP: Make xx the subject of the equation
[−1 point ⇒ 4 / 6 points left]

Isolate xx in the second equation, which can be done in one step:

x6y=36x=6y+36
x6yx=36=6y+36

STEP: Substitute the result into the other equation to find yy
[−2 points ⇒ 2 / 6 points left]

Now we can substitute the expression 6y+366y+36 into the first equation and solve:

3x+4y=23(6y+36)+4y=218y+108+4y=222y=110y=5
3x+4y3(6y+36)+4y18y+108+4y22yy=2=2=2=110=5

STEP: Use the yy-value to find xx
[−2 points ⇒ 0 / 6 points left]

Right now we have only half of the answer! To get the other half, we need to find the value of xx. We can use the equation we got when we isolated xx: it is the most convenient choice because xx is the subject of the equation.

x=6y+36x=6(5)+36x=6
xxx=6y+36=6(5)+36=6

The correct answers are x=6x=6 and y=5y=5.


Submit your answer as: and

2. Linear & quadratic equations

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=2x2+10x+8y=2x8
yy=2x2+10x+8=2x8
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=2x2+10x+8y=2x2+10x+8 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=2x2+10x+8y=2x2+10x+8.

To plot this graph, we need to find the coordinates of the points where the graph will cross the xx- and yy-axes. These points are the xx- and yy-intercepts, respectively. We will then use these points to plot the graph.

The yy-intercept is the quantity in the equation which is not multiplied by xx. For this equation, the yy-intercept is 88. The coordinates of this point are (0;8)(0;8).

The xx-intercepts are the roots of the equation 2x2+10x+8=02x2+10x+8=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1010. This is the coefficient of the middle term. The product of the two numbers must be 1616. This is the product of the coefficient of the first term 22 and the value of the last term 88. The two numbers are 22 and 88. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2+10x+80=2x2+2x+8x+8
00=2x2+10x+8=2x2+2x+8x+8

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=2x(x+1)+8(x+1)0=(2x+8)(x+1)
00=2x(x+1)+8(x+1)=(2x+8)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1x=1 and x=4x=4. The coordinates of these roots are (1;0)(1;0) and (4;0)(4;0).

Using these xx- and yy-intercepts we will draw the graph:


STEP: Plot the graph of y=2x8y=2x8 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the yy-intercept. After that we will calculate root for the linear equation. We remember that the yy-intercept is the quantity in the equation which is not multiplied by xx.

The yy-intercept is 88. Its coordinates are: (0;8)(0;8).

We will work out the root of the equation 0=2x80=2x8.

0=2x8=02x=8x=4
02xx=2x8=8=4=0

The root is: x=4x=4. It's coordinates are: (4;0)(4;0).

We will now plot the equation y=2x8y=2x8 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (2;4)(2;4) and (4;0)(4;0).

We have used a graphical method to determine the solutions of the simultaneous equations: y=2x2+10x+8y=2x2+10x+8 and y=2x8y=2x8. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (2;4)(2;4) and (4;0)(4;0).


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Simultaneous equations: solving by substitution

Solve this pair of simultaneous equations. You should get two coordinate pairs, (x;y)(x;y), for your answers.

y6=7x5x+y=x2+6
y65x+y=7x=x2+6
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate yy.

y6=7xy=7x+6
y6y=7x=7x+6

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (7x+6)(7x+6), into the quadratic equation in place of yy. Be sure to use brackets! Then simplify the equation to get standard form.

5x+y=x2+65x+(7x+6)=x2+62x+6=x2+6x2+2x=0move the terms onto the left sideto keep the quadratic term positive
5x+y5x+(7x+6)2x+6x2+2x=x2+6=x2+6=x2+6=0move the terms onto the left sideto keep the quadratic term positive

STEP: Solve for the values of xx
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two xx-values.

x2+2x=0x(x+2)=0x=2 and x=0
x2+2xx(x+2)x=2=0=0 and x=0

STEP: Find the value of yy for each of the xx-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of yy for these equations. We must substitute in the values x=2x=2 and x=0x=0. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=2:y=7(2)+6y=8If x=0:y=7(0)+6y=6
If x=2:yyIf x=0:yy=7(2)+6=8=7(0)+6=6

Finally we have the answers: if x=2x=2 then y=8y=8 and if x=0x=0 then y=6y=6.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (2;8)(2;8) and (0;6)(0;6).


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Simultaneous equations

Solve for xx and yy:

y=x2+5x+6y=x+3
yy=x2+5x+6=x+3
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+3x+3 in place of yy in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+5x+6linear equationy=x+3
quadratic equationlinear equationy=x2+5x+6y=x+3

This means we want two pairs of xx and yy values which solve both equations.

We will substitute the value of yy from the linear equation into the quadratic equation. Then we will simplify.

x+3=x2+5x+60=x2+5x+6x30=x2+5xx+630=x2+4x+3
x+3000=x2+5x+6=x2+5x+6x3=x2+5xx+63=x2+4x+3

STEP: Solve the equation x2+4x+3=0x2+4x+3=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 44. This is the coefficient of the middle term. The product of the two numbers must be 33. This is the product of the coefficient of the first term 11 and the value of the last term 33. The two numbers are 11 and 33. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+4x+30=x2+x+3x+3
00=x2+4x+3=x2+x+3x+3

We now group the first two terms together and group the last terms together and factorise.

0=x2+x+3x+30=x(x+1)+3(x+1)
00=x2+x+3x+3=x(x+1)+3(x+1)

The last thing now is to factor the highest common factor.

0=(x+3)(x+1)
0=(x+3)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the xx-values of our solutions. The xx-values are 11 and 33.


STEP: Calculate the corresponding values of yy
[−2 points ⇒ 0 / 6 points left]

We will substitute each xx-value into the linear equation. This will give us the corresponding values of yy.

For x=1x=1:

y=x+3=2
y=x+3=2

For x=3x=3:

y=x+3=0
y=x+3=0

The solutions are x=1x=1 together with y=2y=2 , and x=3x=3 with y=0y=0. In coordinates pairs, the solutions are (1;2)(1;2) and (3;0)(3;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;2)(1;2) and (3;0)(3;0). These are the same solutions we have calculated. This means we are 100%100% correct.

The solutions are (1;2)(1;2) and (3;0)(3;0).


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Simultaneous equations

Solve for xx and yy in the following simultaneous equations:

y=3x22x2y=x+2
yy=3x22x2=x+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+2x+2 in place of yy in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x22x2linear equationy=x+2
quadratic equationlinear equationy=3x22x2y=x+2

This means we want two pairs of xx and yy-values which solve both equations.

We will substitute the value of yy from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of yy from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x+2=3x22x20=3x22x2+x20=3x22x+x220=3x2x4
x+2000=3x22x2=3x22x2+x2=3x22x+x22=3x2x4

STEP: Solve the equation 0=3x2x40=3x2x4
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 11. This is the coefficient of the middle term. The product of the two numbers must be 1212. This is the product of the coefficient of the first term 33 and the value of the last term 44. The two numbers are 33 and 44. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2x40=3x2+3x4x4
00=3x2x4=3x2+3x4x4

We now group the first two terms together and group the last terms together and factorise.

0=3x2+3x4x40=3x(x+1)4(x+1)
00=3x2+3x4x4=3x(x+1)4(x+1)

The last thing now is to factor the highest common factor.

0=(3x4)(x+1)
0=(3x4)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the xx-values of our solutions. The xx-values are 4343 and 11.


STEP: Calculate the corresponding values of yy
[−2 points ⇒ 0 / 6 points left]

We will substitute each xx-value into the linear equation. This will give us the corresponding values of yy.

NOTE: We can do that by substituting xx into any of the original equations. It is easier to use the linear one.

For x=43x=43:

y=x+2=23
y=x+2=23

For x=1x=1:

y=x+2=3
y=x+2=3

The solutions are x=43x=43 together with y=23y=23 , and x=1x=1 with y=3y=3. In coordinate pairs, the solutions are (43;23)(43;23) and (1;3)(1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (43;23)(43;23) and (1;3)(1;3). These are the same solutions we have calculated. This means we are 100%100% correct.

The solutions are (43;23)(43;23) and (1;3)(1;3).


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Simultaneous equations

Determine the solution of the following equations:

y=x23x1x=y+4
yx=x23x1=y+4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+4y+4 in place of xx in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x23x1linear equationx=y+4
quadratic equationlinear equationy=x23x1x=y+4

This means we want two pairs of xx and yy values which solve both equations.

We will substitute the value of xx from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of yy from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y+4)23(y+4)10=(y+4)23(y+4)1y0=y2+8y+16(3y+12)1y0=y2+8y+163y121y0=y2+8y3yy+161210=y2+5yy+161210=y2+4y+161210=y2+4y+410=y2+4y+3
y00000000=(y+4)23(y+4)1=(y+4)23(y+4)1y=y2+8y+16(3y+12)1y=y2+8y+163y121y=y2+8y3yy+16121=y2+5yy+16121=y2+4y+16121=y2+4y+41=y2+4y+3

STEP: Solve the equation 0=y2+4y+30=y2+4y+3
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 44. This is the coefficient of the middle term. The product of the two numbers must be 33. This is the product of the coefficient of the first term 11 and the value of the last term 33. The two numbers are 11 and 33. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2+4y+30=y2+y+3y+3
00=y2+4y+3=y2+y+3y+3

We now group the first two terms together and group the last terms together and factorise.

0=y2+y+3y+30=y(y+1)+3(y+1)
00=y2+y+3y+3=y(y+1)+3(y+1)

The last thing now is to factor the highest common factor.

0=(y+3)(y+1)
0=(y+3)(y+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the yy-values of our solutions. The yy-values are -1 and -3.


STEP: Calculate the corresponding values of xx
[−2 points ⇒ 0 / 6 points left]

We will substitute each yy-value into the linear equation. This will give us the corresponding values of xx.

NOTE: We can do that by substituting yy into any of the original equations. It is easier to use the linear one.

For y=1y=1:

x=y+4=3
x=y+4=3

For y=3y=3:

x=y+4=1
x=y+4=1

The solutions are x=3x=3 together with y=1y=1 , and x=1x=1 with y=3y=3. In coordinates pairs, the solutions are (3;1)(3;1) and (1;3)(1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;1)(3;1) and (1;3)(1;3). These are the same solutions we have calculated. This means we are 100%100% correct.

The solutions are (3;1)(3;1) and (1;3)(1;3).


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Simultaneous equations: intersection points

The graph below shows the equations:

y=x32andy=x22+2x32
y=x32andy=x22+2x32

Find the intersection points of the curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation yy is equal to the quantity x32x32; and at the same time, the quadratic equation also contains the value yy. Substitute like this:

y is x32and:yis also x22+2x32Therefore: x32=x22+2x32
yand:yTherefore: x32 is x32is also x22+2x32=x22+2x32

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 22. So multiply both sides of the equation by 22 in order to cancel the denominators. Then arrange the equation in standard form.

(2)(x32)=(x22+2x32)(2)2x3=x2+4x30=x2+6x
(2)(x32)2x30=(x22+2x32)(2)=x2+4x3=x2+6x

Remember that if the quadratic coefficient is negative, we can multiply the equation by −1−1. That will make it easier to factorise.

0=x26x
0=x26x

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two xx-values.

0=x26x0=x(x6)x=0 and x=6
00x=0=x26x=x(x6) and x=6

STEP: Use the values of xx to find the values of yy
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of yy for these equations. Substitute in the values x=0x=0 and x=6x=6 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=0:y=(0)32y=32If x=6:y=(6)32y=152
If x=0:yyIf x=6:yy=(0)32=32=(6)32=152

Finally! The solutions are (0;32)(0;32) and (6;152)(6;152).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (0;32)(0;32) and (6;152)(6;152).


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Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x25x+3x=y3
yx=x25x+3=y3

Which of the following equations shows the correct substitution step?

A y=(y+3)25(y+3)+3y=(y+3)25(y+3)+3
B x=y25y+3x=y25y+3
C y=(y3)25(y3)+3y=(y3)25(y3)+3
D y=y25y+3y=y25y+3
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute y3y3 in place of xx in the quadratic equation y=x25x+3y=x25x+3.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x25x+3linear equationx=y3
quadratic equationlinear equationy=x25x+3x=y3

We need to substitute y3y3 from the linear equation in place of xx in the quadratic equation.

y=(y3)25(y3)+3
y=(y3)25(y3)+3
NOTE: We could also substitute the value of yy from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for yy. The question does not ask us to solve the simultaneous equations.

The correct substitution step is y=(y3)25(y3)+3y=(y3)25(y3)+3 which is choice C.


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Simultaneous equations: special outcomes

Solve for xx and yy in the following simultaneous equations:

y=2x2x1y=3x3
yy=2x2x1=3x3
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 3x33x3 in place of yy in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=2x2x1y=2x2x1 and the linear equation is y=3x3y=3x3. This means we expect two pairs of xx and yy values which solve both equations.

We will substitute the value of yy from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of yy from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
3x3=2x2x10=2x2x13x+30=2x2x3x1+30=2x24x+2
3x3000=2x2x1=2x2x13x+3=2x2x3x1+3=2x24x+2

STEP: Solve the equation 0=2x24x+20=2x24x+2
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 44. This is the coefficient of the middle term. The product of the two numbers must be 44. This is the product of the coefficient of the first term 22 and the value of the last term 22. If we take 22 which is half of the value of 44 (the coefficient of the middle term) and square it, we get the product 44. In this case we will use only one number to factorise, that is, 22. We will re-write the middle term, forming two terms, using this value.

0=2x24x+20=2x22x2x+2
00=2x24x+2=2x22x2x+2

We now group the first two terms together and group the last terms together and factorise.

0=2x22x2x+20=2x(x1)2(x1)
00=2x22x2x+2=2x(x1)2(x1)

The last thing now is to factor the highest common factor.

0=(2x2)(x1)
0=(2x2)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 11. This root is the xx-value of our solution. This xx-value will have one corresponding yy value.


STEP: Calculate the corresponding value of yy
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of xx, we will substitute it into the linear equation to calculate the corresponding value of yy.

NOTE: We can do that by substituting xx into any of the original equations. It is easier to use the linear one.
y=3x3=3(1)3=0
y=3x3=3(1)3=0

The solution is x=1x=1 together with y=0y=0. Using coordinate pairs, the solution is (1;0)(1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;0)(1;0). This is the same solution we have calculated. This means we are 100%100% correct.

The solution is (1;0)(1;0).


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Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for xx and yy:

x=2yandx2+3xy=40
x=2yandx2+3xy=40
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2yx=2y into x2+3xy=40x2+3xy=40 and solve for yy
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for xx and yy:

x=2yandx2+3xy=40
x=2yandx2+3xy=40

First, we can substitute 2y2y in place of xx in the second equation:

(2y)2+3(2y)(y)=40
(2y)2+3(2y)(y)=40

We can simplify this equation to solve for yy:

(2y)2+3(2y)(y)=404y2+6y2=4010y2=40y2=4
(2y)2+3(2y)(y)4y2+6y210y2y2=40=40=40=4

yy can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2
y=2 or y=2

STEP: Use the yy-values to find the xx-values
[−1 point ⇒ 0 / 4 points left]

Each of the yy-values that we have found will have a corresponding value of xx, which can be found from the equation x=2yx=2y.

For y=2y=2:

x=2y=2(2)=4
x=2y=2(2)=4

And for y=2y=2:

x=2y=2(2)=4
x=2y=2(2)=4

So the xx-values are x=4 or x=4x=4 or x=4.

Therefore the correct coordinate pairs are (4;2)(4;2) and (4;2)(4;2).


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Simultaneous equations with xyxy terms

Determine the solution of the following equations:

2y=5xxy5y=x1
2yy=5xxy5=x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1x1 in place of yy in the hyperbolic equation and then solve for xx.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=5xxy5linear equationy=x1
hyperbolic equationlinear equation2y=5xxy5y=x1

This means we want two pairs of xx and yy-values which solve both equations.

We will substitute the value of yy from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either xx or yy from 2y=5xxy52y=5xxy5 and substitute its value into y=x1y=x1. We could still get the same answer, but the calculations might become too long.
2(x1)=5xx(x1)50=5xx(x1)5+2(x1)0=5x(x2x)5+(2x2)0=x24x2x520=x26x520=x26x7
2(x1)00000=5xx(x1)5=5xx(x1)5+2(x1)=5x(x2x)5+(2x2)=x24x2x52=x26x52=x26x7

STEP: Solve the equation 0=x26x70=x26x7
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x26x70=(x7)(x+1)
00=x26x7=(x7)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the xx-values of our solutions. The xx-values are 77 and 11.


STEP: Calculate the corresponding values of yy
[−2 points ⇒ 0 / 6 points left]

By using the values of xx, we will calculate the corresponding values of yy. We will do so by substituting each xx-value into the linear equation.

NOTE: We can still get the same values of yy if we substitute the values xx into the equation 2y=5xxy52y=5xxy5. If we do that, the calculations might become too long.

For x=7x=7:

y=x1=8
y=x1=8

For x=1x=1:

y=x1=0
y=x1=0

The solutions are x=7x=7 together with y=8y=8 , and x=1x=1 with y=0y=0. Using coordinate pairs, the solutions are (7;8)(7;8) and (1;0)(1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (7;8)(7;8) and (1;0)(1;0). These are the same solutions we have calculated. This means we are 100%100% correct.

The solutions are (7;8)(7;8) and (1;0)(1;0).


Submit your answer as: and

Exercises

Points of intersection

Graph the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+3x4y=x7
yy=x2+3x4=x7
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+3x4y=x2+3x4 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+3x4y=x2+3x4.

To plot this graph, we need to find the coordinates of the points where the graph will cross the xx- and yy-axes. These points are the xx- and yy-intercepts, respectively. We will then use these points to plot the graph.

The yy-intercept is the quantity in the equation which is not multiplied by xx. For this equation, the yy-intercept is −4−4. The coordinates of this point are (0;4)(0;4).

The xx-intercepts are the roots of the equation x2+3x4=0x2+3x4=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 33. This is the coefficient of the middle term. The product of the two numbers must be 44. This is the product of the coefficient of the first term 11 and the value of the last term 44. The two numbers are 44 and 11. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+3x40=x2+4xx4
00=x2+3x4=x2+4xx4

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+4)(x+4)0=(x1)(x+4)
00=x(x+4)(x+4)=(x1)(x+4)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1x=1 and x=4x=4. The coordinates of these roots are (1;0)(1;0) and (4;0)(4;0).

Using these xx- and yy-intercepts we will draw the graph:


STEP: Plot the graph of y=x7y=x7 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the yy-intercept. After that we will calculate root for the linear equation. We remember that the yy-intercept is the quantity in the equation which is not multiplied by xx.

The yy-intercept is 77. Its coordinates are: (0;7)(0;7).

We will work out the root of the equation 0=x70=x7.

0=x7=0x=7x=7
0xx=x7=7=7=0

The root is: x=7x=7. It's coordinates are: (7;0)(7;0).

We will now plot the equation y=x7y=x7 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;6)(1;6) and (3;4)(3;4).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+3x4y=x2+3x4 and y=x7y=x7. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;6)(1;6) and (3;4)(3;4).


Submit your answer as: and

Points of intersection

Plot the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+6x+8y=x+2
yy=x2+6x+8=x+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+6x+8y=x2+6x+8 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+6x+8y=x2+6x+8.

To plot this graph, we need to find the coordinates of the points where the graph will cross the xx- and yy-axes. These points are the xx- and yy-intercepts, respectively. We will then use these points to plot the graph.

The yy-intercept is the quantity in the equation which is not multiplied by xx. For this equation, the yy-intercept is 88. The coordinates of this point are (0;8)(0;8).

The xx-intercepts are the roots of the equation x2+6x+8=0x2+6x+8=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 66. This is the coefficient of the middle term. The product of the two numbers must be 88. This is the product of the coefficient of the first term 11 and the value of the last term 88. The two numbers are 22 and 44. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+6x+80=x2+2x+4x+8
00=x2+6x+8=x2+2x+4x+8

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+2)+4(x+2)0=(x+4)(x+2)
00=x(x+2)+4(x+2)=(x+4)(x+2)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=2x=2 and x=4x=4. The coordinates of these roots are (2;0)(2;0) and (4;0)(4;0).

Using these xx- and yy-intercepts we will draw the graph:


STEP: Plot the graph of y=x+2y=x+2 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the yy-intercept. After that we will calculate root for the linear equation. We remember that the yy-intercept is the quantity in the equation which is not multiplied by xx.

The yy-intercept is 22. Its coordinates are: (0;2)(0;2).

We will work out the root of the equation 0=x+20=x+2.

0=x+2=0x=2x=2
0xx=x+2=2=2=0

The root is: x=2x=2. It's coordinates are: (2;0)(2;0).

We will now plot the equation y=x+2y=x+2 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;3)(1;3) and (6;8)(6;8).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+6x+8y=x2+6x+8 and y=x+2y=x+2. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;3)(1;3) and (6;8)(6;8).


Submit your answer as: and

Points of intersection

Sketch the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x27x+6y=x1
yy=x27x+6=x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1)(x1;y1) and (x2;y2)(x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x27x+6y=x27x+6 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x27x+6y=x27x+6.

To plot this graph, we need to find the coordinates of the points where the graph will cross the xx- and yy-axes. These points are the xx- and yy-intercepts, respectively. We will then use these points to plot the graph.

The yy-intercept is the quantity in the equation which is not multiplied by xx. For this equation, the yy-intercept is 66. The coordinates of this point are (0;6)(0;6).

The xx-intercepts are the roots of the equation x27x+6=0x27x+6=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 77. This is the coefficient of the middle term. The product of the two numbers must be 66. This is the product of the coefficient of the first term 11 and the value of the last term 66. The two numbers are 11 and 66. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x27x+60=x2x6x+6
00=x27x+6=x2x6x+6

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x1)6(x1)0=(x6)(x1)
00=x(x1)6(x1)=(x6)(x1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=6x=6 and x=1x=1. The coordinates of these roots are (6;0)(6;0) and (1;0)(1;0).

Using these xx- and yy-intercepts we will draw the graph:


STEP: Plot the graph of y=x1y=x1 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the yy-intercept. After that we will calculate root for the linear equation. We remember that the yy-intercept is the quantity in the equation which is not multiplied by xx.

The yy-intercept is 11. Its coordinates are: (0;1)(0;1).

We will work out the root of the equation 0=x10=x1.

0=x1=0x=1
0x=x1=1=0

The root is: x=1x=1. It's coordinates are: (1;0)(1;0).

We will now plot the equation y=x1y=x1 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (7;6)(7;6) and (1;0)(1;0).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x27x+6y=x27x+6 and y=x1y=x1. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (7;6)(7;6) and (1;0)(1;0).


Submit your answer as: and

Simultaneous equations: solving by substitution

Find the complete solution for these simultaneous equations. You should get two coordinate pairs, (x;y)(x;y), for your answers.

8y=xxy2=4y+3
8yxy2=x=4y+3
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate xx.

8y=xx=8y
8yx=x=8y

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (8y)(8y), into the quadratic equation in place of xx. Be sure to use brackets! Then simplify the equation to get standard form.

xy2=4y+3(8y)y2=4y+3y28y=4y+30=y2+4y+3move the terms onto the right sideto keep the quadratic term positive
xy2(8y)y2y28y0=4y+3=4y+3=4y+3=y2+4y+3move the terms onto the right sideto keep the quadratic term positive

STEP: Solve for the values of yy
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two yy-values.

0=y2+4y+30=(y+1)(y+3)y=3 and y=1
00y=3=y2+4y+3=(y+1)(y+3) and y=1

STEP: Find the value of xx for each of the yy-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of xx for these equations. We must substitute in the values y=3y=3 and y=1y=1. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=3:x=8(3)x=24If y=1:x=8(1)x=8
If y=3:xxIf y=1:xx=8(3)=24=8(1)=8

Finally we have the answers: if x=24x=24 then y=3y=3 and if x=8x=8 then y=1.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (24;3) and (8;1).


Submit your answer as: and

Simultaneous equations: solving by substitution

Solve simultaneously for x and y. You should get two coordinate pairs, (x;y), for your answers.

0=x8y+3y23=x4y
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

0=x8y+3x=8y+3

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (8y+3), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

y23=x4yy23=(8y+3)4yy23=4y3y24y=0move the terms onto the left sideto keep the quadratic term positive

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

y24y=0y(y4)=0y=4 and y=0

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=4 and y=0. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=4:x=8(4)+3x=29If y=0:x=8(0)+3x=3

Finally we have the answers: if x=29 then y=4 and if x=3 then y=0.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (29;4) and (3;0).


Submit your answer as: and

Simultaneous equations: solving by substitution

Solve this pair of simultaneous equations. You should get two coordinate pairs, (x;y), for your answers.

x+2y+9=0xy2=9
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

x+2y+9=0x=2y9

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (2y9), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

xy2=9(2y9)y2=9y22y9=90=y2+2ymove the terms onto the right sideto keep the quadratic term positive

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

0=y2+2y0=y(y+2)y=2 and y=0

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=2 and y=0. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=2:x=2(2)9x=5If y=0:x=2(0)9x=9

Finally we have the answers: if x=5 then y=2 and if x=9 then y=0.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (5;2) and (9;0).


Submit your answer as: and

Simultaneous equations

Solve for x and y:

y=x22x1y=2x4
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x4 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x22x1linear equationy=2x4

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

2x4=x22x10=x22x12x+40=x22x2x1+40=x24x+3

STEP: Solve the equation x24x+3=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x24x+30=x2x3x+3

We now group the first two terms together and group the last terms together and factorise.

0=x2x3x+30=x(x1)3(x1)

The last thing now is to factor the highest common factor.

0=(x3)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 3 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=3:

y=2x4=2

For x=1:

y=2x4=2

The solutions are x=3 together with y=2 , and x=1 with y=2. In coordinates pairs, the solutions are (3;2) and (1;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;2) and (1;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;2) and (1;2).


Submit your answer as: and

Simultaneous equations

Solve the following equations simultaneously:

y=x23x+2y=x1
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x23x+2linear equationy=x1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x1=x23x+20=x23x+2x+10=x23xx+2+10=x24x+3

STEP: Solve the equation x24x+3=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x24x+30=x2x3x+3

We now group the first two terms together and group the last terms together and factorise.

0=x2x3x+30=x(x1)3(x1)

The last thing now is to factor the highest common factor.

0=(x3)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 3 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=3:

y=x1=2

For x=1:

y=x1=0

The solutions are x=3 together with y=2 , and x=1 with y=0. In coordinates pairs, the solutions are (3;2) and (1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;2) and (1;0). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;2) and (1;0).


Submit your answer as: and

Simultaneous equations

Solve for x and y from the given equations:

y=x2+x3y=2x1
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+x3linear equationy=2x1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

2x1=x2+x30=x2+x32x+10=x2+x2x3+10=x2x2

STEP: Solve the equation x2x2=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 2. This is the product of the coefficient of the first term 1 and the value of the last term 2. The two numbers are 1 and 2. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2x20=x2+x2x2

We now group the first two terms together and group the last terms together and factorise.

0=x2+x2x20=x(x+1)2(x+1)

The last thing now is to factor the highest common factor.

0=(x2)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 2 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=2:

y=2x1=3

For x=1:

y=2x1=3

The solutions are x=2 together with y=3 , and x=1 with y=3. In coordinates pairs, the solutions are (2;3) and (1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;3) and (1;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (2;3) and (1;3).


Submit your answer as: and

Simultaneous equations

Solve for x and y:

y=3x2+9x+3y=2x3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x2+9x+3linear equationy=2x3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x3=3x2+9x+30=3x2+9x+3+2x+30=3x2+9x+2x+3+30=3x2+11x+6

STEP: Solve the equation 0=3x2+11x+6
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 11. This is the coefficient of the middle term. The product of the two numbers must be 18. This is the product of the coefficient of the first term 3 and the value of the last term 6. The two numbers are 2 and 9. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2+11x+60=3x2+2x+9x+6

We now group the first two terms together and group the last terms together and factorise.

0=3x2+2x+9x+60=x(3x+2)+3(3x+2)

The last thing now is to factor the highest common factor.

0=(x+3)(3x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 23 and 3.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=23:

y=2x3=53

For x=3:

y=2x3=3

The solutions are x=23 together with y=53 , and x=3 with y=3. In coordinate pairs, the solutions are (23;53) and (3;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (23;53) and (3;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (23;53) and (3;3).


Submit your answer as: and

Simultaneous equations

Determine the solution of the following equations:

y=3x2+7x+1y=2x+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x+3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x2+7x+1linear equationy=2x+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x+3=3x2+7x+10=3x2+7x+12x30=3x2+7x2x+130=3x2+5x2

STEP: Solve the equation 0=3x2+5x2
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 3 and the value of the last term 2. The two numbers are 6 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2+5x20=3x2+6xx2

We now group the first two terms together and group the last terms together and factorise.

0=3x2+6xx20=3x(x+2)(x+2)

The last thing now is to factor the highest common factor.

0=(3x1)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 13 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=13:

y=2x+3=113

For x=2:

y=2x+3=1

The solutions are x=13 together with y=113 , and x=2 with y=1. In coordinate pairs, the solutions are (13;113) and (2;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (13;113) and (2;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (13;113) and (2;1).


Submit your answer as: and

Simultaneous equations

Solve for x and y from the given equations:

y=2x28x+6y=x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x28x+6linear equationy=x1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x1=2x28x+60=2x28x+6x+10=2x28xx+6+10=2x29x+7

STEP: Solve the equation 0=2x29x+7
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 9. This is the coefficient of the middle term. The product of the two numbers must be 14. This is the product of the coefficient of the first term 2 and the value of the last term 7. The two numbers are 2 and 7. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x29x+70=2x22x7x+7

We now group the first two terms together and group the last terms together and factorise.

0=2x22x7x+70=2x(x1)7(x1)

The last thing now is to factor the highest common factor.

0=(2x7)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 72 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=72:

y=x1=52

For x=1:

y=x1=0

The solutions are x=72 together with y=52 , and x=1 with y=0. In coordinate pairs, the solutions are (72;52) and (1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (72;52) and (1;0). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (72;52) and (1;0).


Submit your answer as: and

Simultaneous equations

Solve the following equations simultaneously:

y=x22x2x=y+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+2 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x22x2linear equationx=y+2

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y+2)22(y+2)20=(y+2)22(y+2)2y0=y2+4y+4(2y+4)2y0=y2+4y+42y42y0=y2+4y2yy+4420=y2+2yy+4420=y2+y+4420=y2+y2

STEP: Solve the equation 0=y2+y2
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 2. This is the product of the coefficient of the first term 1 and the value of the last term 2. The two numbers are 2 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2+y20=y2+2yy2

We now group the first two terms together and group the last terms together and factorise.

0=y2+2yy20=y(y+2)(y+2)

The last thing now is to factor the highest common factor.

0=(y1)(y+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 1 and -2.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=1:

x=y+2=3

For y=2:

x=y+2=0

The solutions are x=3 together with y=1 , and x=0 with y=2. In coordinates pairs, the solutions are (3;1) and (0;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;1) and (0;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;1) and (0;2).


Submit your answer as: and

Simultaneous equations

Solve for x and y from the given equations:

y=x22x1x=y+1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+1 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x22x1linear equationx=y+1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y+1)22(y+1)10=(y+1)22(y+1)1y0=y2+2y+1(2y+2)1y0=y2+2y+12y21y0=y2+2y2yy+1210=y2y+1210=y2y110=y2y2

STEP: Solve the equation 0=y2y2
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 2. This is the product of the coefficient of the first term 1 and the value of the last term 2. The two numbers are 1 and 2. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2y20=y2+y2y2

We now group the first two terms together and group the last terms together and factorise.

0=y2+y2y20=y(y+1)2(y+1)

The last thing now is to factor the highest common factor.

0=(y2)(y+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 2 and -1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=2:

x=y+1=3

For y=1:

x=y+1=0

The solutions are x=3 together with y=2 , and x=0 with y=1. In coordinates pairs, the solutions are (3;2) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;2) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;2) and (0;1).


Submit your answer as: and

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=x2+3x+4x=y4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y4 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+3x+4linear equationx=y4

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y4)2+3(y4)+40=(y4)2+3(y4)+4y0=y28y+16+(3y12)+4y0=y28y+16+3y12+4y0=y28y+3yy+1612+40=y25yy+1612+40=y26y+1612+40=y26y+4+40=y26y+8

STEP: Solve the equation 0=y26y+8
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 8. This is the product of the coefficient of the first term 1 and the value of the last term 8. The two numbers are 2 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y26y+80=y22y4y+8

We now group the first two terms together and group the last terms together and factorise.

0=y22y4y+80=y(y+2)+4(y+2)

The last thing now is to factor the highest common factor.

0=(y+4)(y+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 4 and 2.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=4:

x=y4=0

For y=2:

x=y4=2

The solutions are x=0 together with y=4 , and x=2 with y=2. In coordinates pairs, the solutions are (0;4) and (2;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (0;4) and (2;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (0;4) and (2;2).


Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

y=9x2+152andy=3x229x26

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 9x2+152; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 9x2+152and:yis also 3x229x26Therefore: 9x2+152=3x229x26

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 2. So multiply both sides of the equation by 2 in order to cancel the denominators. Then arrange the equation in standard form.

(2)(9x2+152)=(3x229x26)(2)9x+15=3x29x120=3x227

In this case, there is a common factor of 3 for all of the terms. Divide out this factor before factorising.

0=x29

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x290=(x3)(x+3)x=3 and x=3

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=3 and x=3 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=3:y=9(3)2+152y=21If x=3:y=9(3)2+152y=6

Finally! The solutions are (3;21) and (3;6).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (3;21) and (3;6).


Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

y=3x4+214andy=3x24+21x4+9

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 3x4+214; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 3x4+214and:yis also 3x24+21x4+9Therefore: 3x4+214=3x24+21x4+9

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 4. So multiply both sides of the equation by 4 in order to cancel the denominators. Then arrange the equation in standard form.

(4)(3x4+214)=(3x24+21x4+9)(4)3x+21=3x2+21x+360=3x2+18x+15

In this case, there is a common factor of 3 for all of the terms. Divide out this factor before factorising.

0=x2+6x+5

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2+6x+50=(x+1)(x+5)x=5 and x=1

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=5 and x=1 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=5:y=3(5)4+214y=32If x=1:y=3(1)4+214y=92

Finally! The solutions are (5;32) and (1;92).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (5;32) and (1;92).


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Simultaneous equations: intersection points

The graph below shows the equations:

y=x31andy=x233

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity x31; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is x31and:yis also x233Therefore: x31=x233

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(x31)=(x233)(3)x3=x290=x2+x6


STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2+x60=(x2)(x+3)x=3 and x=2

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=3 and x=2 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=3:y=(3)31y=0If x=2:y=(2)31y=53

Finally! The solutions are (3;0) and (2;53).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (3;0) and (2;53).


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Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2+x3y=x+5

Which of the following equations shows the correct substitution step?

A x+5=x2+x3
B x5=x2+x3
C x=x2+x3
D x+5=y2+y3
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute x+5 in place of y in the quadratic equation y=x2+x3.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2+x3linear equationy=x+5

We need to substitute x+5 from the linear equation in place of y in the quadratic equation.

x+5=x2+x3
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is x+5=x2+x3 which is choice A.


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Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x22x4y=x+2

Which of the following equations shows the correct substitution step?

A x+2=y22y4
B x2=x22x4
C x+2=x22x4
D y=(x+2)22(x+2)4
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute x+2 in place of y in the quadratic equation y=x22x4.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x22x4linear equationy=x+2

We need to substitute x+2 from the linear equation in place of y in the quadratic equation.

x+2=x22x4
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is x+2=x22x4 which is choice C.


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Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x24x+1x=y+5

Which of the following equations shows the correct substitution step?

A x=y24y+1
B y=y24y+1
C y+5=y24y
D y=(y+5)24(y+5)+1
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute y+5 in place of x in the quadratic equation y=x24x+1.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x24x+1linear equationx=y+5

We need to substitute y+5 from the linear equation in place of x in the quadratic equation.

y=(y+5)24(y+5)+1
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for y. The question does not ask us to solve the simultaneous equations.

The correct substitution step is y=(y+5)24(y+5)+1 which is choice D.


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Simultaneous equations: special outcomes

Solve for x and y from the given equations:

y=3x2+2x+1y=2x5
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting 2x5 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=3x2+2x+1 and the linear equation is y=2x5. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x5=3x2+2x+10=3x2+2x+1+2x+50=3x2+2x+2x+1+50=3x2+4x+6

STEP: Solve the equation 0=3x2+4x+6
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 18. This is the product of the coefficient of the first term 3 and the value of the last term 6. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form ax2+bx+c=0:

x=b±(b24ac)2a

Before we can use the quadratic formula, we need to identify the coefficients a, b and c from the equation 3x2+4x+6=0 These are:

a=3,b=4 and c=6.

We will substitute these values into the formula and simplify.

x=(4)±((4)24(3)(6))2(3)=4±(16(72))6=4±(56)6

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that y=3x2+2x+1 and y=2x5 cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.


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Simultaneous equations: special outcomes

Solve the following equations simultaneously:

y=x2x+2y=3x2
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 3x2 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=x2x+2 and the linear equation is y=3x2. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
3x2=x2x+20=x2x+23x+20=x2x3x+2+20=x24x+4

STEP: Solve the equation 0=x24x+4
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. If we take 2 which is half of the value of 4 (the coefficient of the middle term) and square it, we get the product 4. In this case we will use only one number to factorise, that is, 2. We will re-write the middle term, forming two terms, using this value.

0=x24x+40=x22x2x+4

We now group the first two terms together and group the last terms together and factorise.

0=x22x2x+40=x(x2)2(x2)

The last thing now is to factor the highest common factor.

0=(x2)(x2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 2. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=3x2=3(2)2=4

The solution is x=2 together with y=4. Using coordinate pairs, the solution is (2;4).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;4). This is the same solution we have calculated. This means we are 100% correct.

The solution is (2;4).


Submit your answer as:

Simultaneous equations: special outcomes

Solve for x and y in the following simultaneous equations:

y=x2+2y=2x+1
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 2x+1 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=x2+2 and the linear equation is y=2x+1. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x+1=x2+20=x2+2+2x10=x2+2x+210=x2+2x+1

STEP: Solve the equation 0=x2+2x+1
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 2. This is the coefficient of the middle term. The product of the two numbers must be 1. This is the product of the coefficient of the first term 1 and the value of the last term 1. If we take 1 which is half of the value of 2 (the coefficient of the middle term) and square it, we get the product 1. In this case we will use only one number to factorise, that is, 1. We will re-write the middle term, forming two terms, using this value.

0=x2+2x+10=x2+x+x+1

We now group the first two terms together and group the last terms together and factorise.

0=x2+x+x+10=x(x+1)+(x+1)

The last thing now is to factor the highest common factor.

0=(x+1)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 1. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=2x+1=2(1)+1=3

The solution is x=1 together with y=3. Using coordinate pairs, the solution is (1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;3). This is the same solution we have calculated. This means we are 100% correct.

The solution is (1;3).


Submit your answer as:

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=3yandx2+5xy=24
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=3y into x2+5xy=24 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=3yandx2+5xy=24

First, we can substitute 3y in place of x in the second equation:

(3y)2+5(3y)(y)=24

We can simplify this equation to solve for y:

(3y)2+5(3y)(y)=249y215y2=246y2=24y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=3y.

For y=2:

x=3y=3(2)=6

And for y=2:

x=3y=3(2)=6

So the x-values are x=6 or x=6.

Therefore the correct coordinate pairs are (6;2) and (6;2).


Submit your answer as: and

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx24xy=16
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x24xy=16 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx24xy=16

First, we can substitute 2y in place of x in the second equation:

(2y)24(2y)(y)=16

We can simplify this equation to solve for y:

(2y)24(2y)(y)=164y28y2=164y2=16y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=2:

x=2y=2(2)=4

And for y=2:

x=2y=2(2)=4

So the x-values are x=4 or x=4.

Therefore the correct coordinate pairs are (4;2) and (4;2).


Submit your answer as: and

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx23xy=32
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x23xy=32 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx23xy=32

First, we can substitute 2y in place of x in the second equation:

(2y)23(2y)(y)=32

We can simplify this equation to solve for y:

(2y)23(2y)(y)=324y26y2=322y2=32y2=16

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=4 or y=4

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=4:

x=2y=2(4)=8

And for y=4:

x=2y=2(4)=8

So the x-values are x=8 or x=8.

Therefore the correct coordinate pairs are (8;4) and (8;4).


Submit your answer as: and

Simultaneous equations with xy terms

Solve for x and y in the following simultaneous equations:

y=5xxy+3x=y+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+3 in place of x in the hyperbolic equation and then solve for y.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equationy=5xxy+3linear equationx=y+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of x from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from y=5xxy+3 and substitute its value into x=y+3. We could still get the same answer, but the calculations might become too long.
y=5(y+3)(y+3)y+30=5(y+3)(y+3)y+3y0=5(y+3)y(y+3)+3y0=y25y3yy+15+30=y28yy+15+30=y29y+15+30=y29y+18

STEP: Solve the equation 0=y29y+18
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y29y+180=(y3)(y6)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 3 and 6.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

By using the values of y, we will calculate the corresponding values of x. We will do so by substituting each y-value into the linear equation.

NOTE: We can still get the same values of x if we substitute the values y into the equation y=5xxy+3. If we do that, the calculations might become too long.

For y=3:

x=y+3=0

For y=6:

x=y+3=3

The solutions are x=0 together with y=3 , and x=3 with y=6. Using coordinate pairs, the solutions are (0;3) and (3;6).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (0;3) and (3;6). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (0;3) and (3;6).


Submit your answer as: and

Simultaneous equations with xy terms

Solve for x and y in the following simultaneous equations:

3y=8xxy3x=y1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y1 in place of x in the hyperbolic equation and then solve for y.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation3y=8xxy3linear equationx=y1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of x from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 3y=8xxy3 and substitute its value into x=y1. We could still get the same answer, but the calculations might become too long.
3y=8(y1)(y1)y30=8(y1)(y1)y33y0=8(y1)y(y1)33y0=y2+8y+y3y+830=y2+9y3y+830=y2+6y+830=y2+6y+5

STEP: Solve the equation 0=y2+6y+5
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2+6y+50=(y+5)(y+1)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 5 and 1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

By using the values of y, we will calculate the corresponding values of x. We will do so by substituting each y-value into the linear equation.

NOTE: We can still get the same values of x if we substitute the values y into the equation 3y=8xxy3. If we do that, the calculations might become too long.

For y=5:

x=y1=4

For y=1:

x=y1=0

The solutions are x=4 together with y=5 , and x=0 with y=1. Using coordinate pairs, the solutions are (4;5) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;5) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;5) and (0;1).


Submit your answer as: and

Simultaneous equations with xy terms

Determine the solution of the following equations:

3y=3xxy+1y=2x3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x3 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation3y=3xxy+1linear equationy=2x3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 3y=3xxy+1 and substitute its value into y=2x3. We could still get the same answer, but the calculations might become too long.
3(2x3)=3xx(2x3)+10=3xx(2x3)+13(2x3)0=3x(2x23x)+1(6x9)0=2x2+6x+6x+1+90=2x2+12x+1+90=2x2+12x+10

STEP: Solve the equation 0=2x2+12x+10
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2+12x+100=(x+1)(x+5)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 5.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation 3y=3xxy+1. If we do that, the calculations might become too long.

For x=1:

y=2x3=1

For x=5:

y=2x3=7

The solutions are x=1 together with y=1 , and x=5 with y=7. Using coordinate pairs, the solutions are (1;1) and (5;7).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;1) and (5;7). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;1) and (5;7).


Submit your answer as: and

3. Practical applications

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 110.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 110.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis110n1×n2=110

The correct equation for the first fact is n1×n2=110.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 110. n1×n2=110
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

Word problems: distance, speed and time

Two missiles start out at two different launchers. The missiles are 1 400 km apart. The missiles start moving towards each other. One missile is moving at 320 km/h and the other missile at 380 km/h.

If both missiles started their journey at the same time, how long will they take to pass each other?

Answer: The two missiles pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two missiles must be equal to the total distance when the missiles meet:

d1+d2=dtotald1+d2=1 400 km

STEP: Write equations to describe the motion of the missiles
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the missiles to meet. Let the time taken be t. Then we can write an expression for the distance each of the missiles travels:

For missile 1:

d1=s1tThe speed of the firstmissile is 320 km/hd1=320t

For missile 2:

d2=s2tThe speed of the secondmissile is 380 km/hd2=380t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=1 400(320t)+(380t)=1 400700t=1 400t=1 400700t=2

The missiles will meet after 2 hours.


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Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 7 sandwiches and 8 pizzas. Here are some facts about their lunch:

  • the total cost for the 7 sandwiches and 8 pizzas is R432
  • a sandwich costs R6 more than a pizza

What is the price for one sandwich and the price for one pizza?

Answer:

A sandwich costs R and a pizza costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a sandwich and a pizza). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

s=the price of a sandwichp=the price of a pizza

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 7 sandwiches and 8 pizzas is R432." We can use the expression 7s to represent the price of the 7 sandwiches. Similarly, the expression 8p represents the price of the 8 pizzas. With these values we can write a full equation for the prices:

In words: the total cost for the 7 sandwiches and 8 pizzas is R432In maths: 7s+8p=432

Now we can use the second point. It says, "a sandwich costs R6 more than a pizza." We can write this as an equation like this:

In words: a sandwich costs R6 more than a pizzaIn maths: s=p+6

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

7s+8p=432s=p+6

We could use elimination, but substitution is a better choice (because s is already the subject). Substitute the second equation into the first equation and solve!

7s+8p=4327(p+6)+8p=4327p+42+8p=43215p=43242p=39015=26

This means that the price of one pizza is R26.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for p to find the value of s (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because s is already isolated).

s=p+6=26+6=32

The price for one sandwich is R32.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is R32 while a pizza costs R26.


Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Adekemi buys some bananas and some chocolates. The total cost for 2 bananas and 3 chocolates is R19,40. And each chocolate costs R2,20 less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like R4,30, or must a price always be a whole number like R4,00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -R5,50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Songezo is drawing a rectangle in his notebook. The length of the rectangle is 3 cm more than its width. The area of the rectangle is 40 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3,5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3,5 days of school, but a rectangle certainly can have a width of 3,5 cm. Or 5,927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: test scores and simultaneous equations

Mpho and Danjuma are friends. Mpho takes Danjuma's civil technology test paper and says: “I have 6 marks more than you do and the sum of both our marks is equal to 152. What are our marks?”

Answer:

Mpho got marks and Danjuma got marks for the civil technology test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

m= Mpho's markd= Danjuma's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

Mpho has 6 marks more than Danjumam=d+6The sum of the marks(the total) is 152m+d=152

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

m+d=152(d+6)+d=1522d=1526d=1462=73

This means that Danjuma's mark is 73.


STEP: Find Mpho's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Danjuma's mark and one of the equations we have to find Mpho's mark. The easiest way to do that is to substitute Danjuma's mark back into the first equation:

m=d+6=(73)+6=79

The students achieved these marks: Mpho earned 79 and Danjuma earned 73.


Submit your answer as: and

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 12.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 12. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 12". Remember that sum means addition. So:

n1+n2=12

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=12, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=122n1+2=122n1=10n1=5

The result is n1=5. Notice that this means that the other number, n2, must be 7, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 12, and 5+7=12.

The smaller number is 5.


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Word problems: checking answers

Jidda is 13 years younger than his sister, Amogelang. In 9 years, Amogelang will be 2 times as old as Jidda. How old is Jidda now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Jidda is years old.

numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Jidda's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Jidda's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Amogelang and Jidda. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aA=Amogelang's ageaJ=Jidda's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Jidda is 13 years younger than his sister, Amogelang." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aJ=aA13

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 9 years, Amogelang will be 2 times as old as Jidda." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aA+9=Amogelang's age in 9 yearsaJ+9=Jidda's age in 9 years

These are the ages at which Amogelang will be 2 times as old as Jidda. We can put all this together as follows:

Amogelang's agein 9 years=2×Jidda's agein 9 yearsaA+9=2(aJ+9)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aA and aJ. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aJ=aA13. We can substitute this into the second equation and solve for aA.

aA+9=2(aJ+9)aA+9=2((aA13)+9)aA+9=2aA817=aA

Terrific: this means that Amogelang is 17 years old. But the question asked for us to find Jidda's age. We can find it using the first equation, which relates the two ages:

aJ=aA13=(17)13=4

So we finally got the answer to the question. Jidda is 4 years old.

The correct answer is: 4.


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Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 15.

Determine the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 15. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 15". Remember that product means multiplication. So:

n1n2=15

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=15, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=15n21+2n1=15n21+2n115=0(n1+5)(n13)=0
n1=5andn1=3

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=3.

This means the other number, n2, must be 5, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 15, and 35=15.

The smaller number is 3.


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Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 90

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 90.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 90. That means if we multiply the numbers we get 90.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 90. In other words, if we multiply the numbers, we get 90. As an equation this is:

n1n2=90

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=90n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=90(n1)2+n190=0(n1+10)(n19)=0
n1=10andn1=9

The solutions to the equation are n1=10 or n1=9. This means the first number is either −10 or 9.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −10 and 9 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=9, so the first number is n1=10.

Based on that we can find the second number. The second number is n1+1, which is equal to (10)+1=9.

The two consecutive integers are −10 and −9.


Submit your answer as: and

Word problems: an age-old question

Zenzile has a son, Chizoba. Here are some facts about how old Zenzile and Chizoba are:

  • Zenzile is 4 times as old as Chizoba right now.
  • 7 years from now, Zenzile will be 3 times as old as Chizoba.

How old are Zenzile and Chizoba now?

Answer:

Zenzile is years old and Chizoba is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Zenzile and his son, Chizoba. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let z=Zenzile's ageLet c=Chizoba's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Zenzile is now 4 times as old as Chizoba." As an equation, this is:

Ages now: z=4c

The second piece of information says that in "7 years... Zenzile will be 3 times as old as his son." In 7 years Zenzile will be z+7 years old, and similarly Chizoba will be c+7 years old. Then:

Ages in 7 years: z+7=3(c+7)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel z.)

z+7=3(c+7)(4c)+7=3c+21c=14

Great! Now we know that Chizoba is 14 years old.


STEP: Use Chizoba's age to find Zenzile's age
[−1 point ⇒ 0 / 6 points left]

We can now find Zenzile's age. Substitute Chizoba's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

z=4c=4(14)=56

Write your final answer: Zenzile is 56 years old and Chizoba is 14 years old.


Submit your answer as: and

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "One Metre at a Time" marathon in Kimberley. The second-place runner finished 13,0 seconds after the winner. The third runner was 19,64 seconds behind the second runner, and the fourth runner was 20,96 seconds behind the third runner. If the second-place runner had an average speed of 19,63 km/h, how many of these runners finished the marathon in less than 2,08 hours?

Can the answer to the question above be a non-integer number?

Answer:

Can the answer be non-integer?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be non-integer
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: how many of these runners finished the marathon in less than 2,08 hours. So the question is about the number of people who finished the race before 2,08 hours had passed. The answer cannot be a non-integer. The number of people to finish before 2,08 hours might be 3 or 7 people, but it cannot be 4,26 people. In other words, the number of people must be a whole number.

The correct answer is: No, it must be an integer.


Submit your answer as:

Word problems: rectangle facts

The diagonal of a rectangle is 28 cm more than its width. The length of the same rectangle is 14 cm more than its width.

What are the width and the length of the rectangle?

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

the diagonal of the rectangle is28 cm more than its widthd=w+28the length of the rectangle is14 cm more than its widthl=w+14

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+28 in for d and w+14 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+28)2=w2+(w+14)2w2+56w+784=w2+(w2+28w+196)0=w228w588

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w228w5880=(w42)(w+14)w=42or w=14

This means that the width of the rectangle is either 42 cm or 14 cm. But the dimensions of a rectangle cannot be negative, so w=42 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+14=(42)+14=56 cm

The width of the rectangle is 42 cm and the length is 56 cm.


Submit your answer as: and

Setting up simultaneous equations

Last week, Abiodun and Bandile had a maths test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 141.
  • Abiodun's mark is 7 more than Bandile's mark.

Let a represent Abiodun's mark and b represent Bandile's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 141.
Abiodun's mark is 7 more than Bandile's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Abiodun's mark and b for Bandile's mark. So we can break up the first fact like this:

The sum of the marksis141a+b=141

The first fact is equivalent to this equation: a+b=141.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Abiodun's markis7 more than Bandile's marka=b+7

This equation, a=b+7, means that Abiodun's mark is more that Bandile's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 141. a+b=141
Abiodun's mark is 7 more than Bandile's mark. a=b+7

Submit your answer as: and

Exercises

Setting up simultaneous equations

Here are facts about two numbers:

  • The sum of the numbers is 17.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The sum of the numbers is 17.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The sum of the numbersis17n1+n2=17

The correct equation for the first fact is n1+n2=17.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The sum of the numbers is 17. n1+n2=17
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 42.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 42.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis42n1×n2=42

The correct equation for the first fact is n1×n2=42.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 42. n1×n2=42
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 132.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 132.
The numbers are consecutive numbers.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis132n1×n2=132

The correct equation for the first fact is n1×n2=132.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 132. n1×n2=132
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

Word problems: distance, speed and time

Two boats start out at two different harbours. The boats are 168 km apart. The boats start moving towards each other. One boat is moving at 82 km/h and the other boat at 86 km/h.

If both boats started their journey at the same time, how long will they take to pass each other?

Answer: The two boats pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two boats must be equal to the total distance when the boats meet:

d1+d2=dtotald1+d2=168 km

STEP: Write equations to describe the motion of the boats
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the boats to meet. Let the time taken be t. Then we can write an expression for the distance each of the boats travels:

For boat 1:

d1=s1tThe speed of the firstboat is 82 km/hd1=82t

For boat 2:

d2=s2tThe speed of the secondboat is 86 km/hd2=86t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=168(82t)+(86t)=168168t=168t=168168t=1

The boats will meet after 1 hours.


Submit your answer as:

Word problems: distance, speed and time

Two trucks start out at two different factories. The trucks are 1 776 km apart. The trucks start travelling towards each other. One truck is travelling at 62 km/h and the other truck at 86 km/h.

If both trucks started their journey at the same time, how long will they take to pass each other?

Answer: The two trucks pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two trucks must be equal to the total distance when the trucks meet:

d1+d2=dtotald1+d2=1 776 km

STEP: Write equations to describe the motion of the trucks
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the trucks to meet. Let the time taken be t. Then we can write an expression for the distance each of the trucks travels:

For truck 1:

d1=s1tThe speed of the firsttruck is 62 km/hd1=62t

For truck 2:

d2=s2tThe speed of the secondtruck is 86 km/hd2=86t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=1 776(62t)+(86t)=1 776148t=1 776t=1 776148t=12

The trucks will meet after 12 hours.


Submit your answer as:

Word problems: distance, speed and time

Two missiles start out at two different launchers. The missiles are 896 km apart. The missiles start moving towards each other. One missile is moving at 172 km/h and the other missile at 276 km/h.

If both missiles started their journey at the same time, how long will they take to pass each other?

Answer: The two missiles pass each other after hours.
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two missiles must be equal to the total distance when the missiles meet:

d1+d2=dtotald1+d2=896 km

STEP: Write equations to describe the motion of the missiles
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the missiles to meet. Let the time taken be t. Then we can write an expression for the distance each of the missiles travels:

For missile 1:

d1=s1tThe speed of the firstmissile is 172 km/hd1=172t

For missile 2:

d2=s2tThe speed of the secondmissile is 276 km/hd2=276t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=896(172t)+(276t)=896448t=896t=896448t=2

The missiles will meet after 2 hours.


Submit your answer as:

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 4 milkshakes and 2 waffles. Here are some facts about their lunch:

  • the total cost for the 4 milkshakes and 2 waffles is R152
  • a milkshake costs R8 more than a waffle

What is the price for one milkshake and the price for one waffle?

Answer:

A milkshake costs R and a waffle costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a milkshake and a waffle). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

m=the price of a milkshakew=the price of a waffle

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 4 milkshakes and 2 waffles is R152." We can use the expression 4m to represent the price of the 4 milkshakes. Similarly, the expression 2w represents the price of the 2 waffles. With these values we can write a full equation for the prices:

In words: the total cost for the 4 milkshakes and 2 waffles is R152In maths: 4m+2w=152

Now we can use the second point. It says, "a milkshake costs R8 more than a waffle." We can write this as an equation like this:

In words: a milkshake costs R8 more than a waffleIn maths: m=w+8

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

4m+2w=152m=w+8

We could use elimination, but substitution is a better choice (because m is already the subject). Substitute the second equation into the first equation and solve!

4m+2w=1524(w+8)+2w=1524w+32+2w=1526w=15232w=1206=20

This means that the price of one waffle is R20.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for w to find the value of m (the price of a milkshake). We can use either equation to do this, but the second one is easier to use (because m is already isolated).

m=w+8=20+8=28

The price for one milkshake is R28.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the milkshake is R28 while a waffle costs R20.


Submit your answer as: and

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 6 wraps and 8 hamburgers. Here are some facts about their lunch:

  • the total cost for the 6 wraps and 8 hamburgers is R430
  • a wrap costs R4 more than a hamburger

What is the price for one wrap and the price for one hamburger?

Answer:

A wrap costs R and a hamburger costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a wrap and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

w=the price of a wraph=the price of a hamburger

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 6 wraps and 8 hamburgers is R430." We can use the expression 6w to represent the price of the 6 wraps. Similarly, the expression 8h represents the price of the 8 hamburgers. With these values we can write a full equation for the prices:

In words: the total cost for the 6 wraps and 8 hamburgers is R430In maths: 6w+8h=430

Now we can use the second point. It says, "a wrap costs R4 more than a hamburger." We can write this as an equation like this:

In words: a wrap costs R4 more than a hamburgerIn maths: w=h+4

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

6w+8h=430w=h+4

We could use elimination, but substitution is a better choice (because w is already the subject). Substitute the second equation into the first equation and solve!

6w+8h=4306(h+4)+8h=4306h+24+8h=43014h=43024h=40614=29

This means that the price of one hamburger is R29.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for h to find the value of w (the price of a wrap). We can use either equation to do this, but the second one is easier to use (because w is already isolated).

w=h+4=29+4=33

The price for one wrap is R33.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the wrap is R33 while a hamburger costs R29.


Submit your answer as: and

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 7 pizzas and 5 sandwiches. Here are some facts about their lunch:

  • the total cost for the 7 pizzas and 5 sandwiches is R337
  • a pizza costs R7 more than a sandwich

What is the price for one pizza and the price for one sandwich?

Answer:

A pizza costs R and a sandwich costs R .

numeric
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a pizza and a sandwich). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

p=the price of a pizzas=the price of a sandwich

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 7 pizzas and 5 sandwiches is R337." We can use the expression 7p to represent the price of the 7 pizzas. Similarly, the expression 5s represents the price of the 5 sandwiches. With these values we can write a full equation for the prices:

In words: the total cost for the 7 pizzas and 5 sandwiches is R337In maths: 7p+5s=337

Now we can use the second point. It says, "a pizza costs R7 more than a sandwich." We can write this as an equation like this:

In words: a pizza costs R7 more than a sandwichIn maths: p=s+7

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

7p+5s=337p=s+7

We could use elimination, but substitution is a better choice (because p is already the subject). Substitute the second equation into the first equation and solve!

7p+5s=3377(s+7)+5s=3377s+49+5s=33712s=33749s=28812=24

This means that the price of one sandwich is R24.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for s to find the value of p (the price of a pizza). We can use either equation to do this, but the second one is easier to use (because p is already isolated).

p=s+7=24+7=31

The price for one pizza is R31.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the pizza is R31 while a sandwich costs R24.


Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are fruits and sweets for sale. Habubakar buys some bananas and some chocolates. The total number of items he buys is 6. Each banana costs R5,70 and each chocolate costs R4,40. The total cost is R30,30. How many bananas did Habubakar buy?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of items be a decimal number (or must it be an integer)?
  2. Can the number of items be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of bananas Habubakar buys. Can the number of items be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the number of items can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the number of items can be a decimal number or not. In other words, can the number of bananas be a decimal number like 3,7 bananas? Or must it be a whole number like 4 bananas? This is different from the airtime example above: airtime can be a decimal number, but it is not possible to buy 3,5 bananas! (You might think, "Wait! I can buy 3,5 bananas, just cut one of them in half." But no one should expect to do this while shopping.)

The answer for the first question is: No, the number of items cannot be a decimal number.


STEP: Decide if the number of items can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the number of items can be negative, or must be postive. The number of bananas cannot be negative. If Habubakar did not buy any bananas at all, then the number of bananas is zero. But is it impossible for him to buy 3 bananas.

The answer to the second question is: No, the number of items cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are fruits and sweets for sale. Chibueze buys some bananas and some chocolates. The total number of items he buys is 8. Each banana costs R4,20 and each chocolate costs R4,80. The total cost is R36,60. How many bananas did Chibueze buy?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of items be a decimal number (or must it be an integer)?
  2. Can the number of items be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of bananas Chibueze buys. Can the number of items be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the number of items can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the number of items can be a decimal number or not. In other words, can the number of bananas be a decimal number like 3,7 bananas? Or must it be a whole number like 4 bananas? This is different from the airtime example above: airtime can be a decimal number, but it is not possible to buy 3,5 bananas! (You might think, "Wait! I can buy 3,5 bananas, just cut one of them in half." But no one should expect to do this while shopping.)

The answer for the first question is: No, the number of items cannot be a decimal number.


STEP: Decide if the number of items can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the number of items can be negative, or must be postive. The number of bananas cannot be negative. If Chibueze did not buy any bananas at all, then the number of bananas is zero. But is it impossible for him to buy 3 bananas.

The answer to the second question is: No, the number of items cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Tanya buys some bananas and some chocolates. The total cost for 2 bananas and 5 chocolates is R33,10. And each chocolate costs R2,00 more than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be R6,75. Or R5. Or R19,87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like R4,30, or must a price always be a whole number like R4,00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -R5,50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Adekunbi is drawing a rectangle in her notebook. The length of the rectangle is 3 cm more than its width. The area of the rectangle is 18 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3,5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3,5 days of school, but a rectangle certainly can have a width of 3,5 cm. Or 5,927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Chichi is drawing a rectangle in her notebook. The length of the rectangle is 5 cm more than its width. The area of the rectangle is 14 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3,5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3,5 days of school, but a rectangle certainly can have a width of 3,5 cm. Or 5,927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Musa is drawing a pattern of rectangles in his notebook. The number of rectangles in each figure is 2 more than the number of rectangles in the previous figure. If there are 7 rectangles in the third figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Musa's diagram. Can the number of rectangles be a decimal, like 4,5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176,3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3,7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3,7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3,7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Musa drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

Word problems: test scores and simultaneous equations

Tshegofatso and Ibironke are friends. Tshegofatso takes Ibironke's civil technology test paper and says: “I have 6 marks more than you do and the sum of both our marks is equal to 162. What are our marks?”

Answer:

Tshegofatso got marks and Ibironke got marks for the civil technology test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

t= Tshegofatso's marki= Ibironke's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

Tshegofatso has 6 marks more than Ibironket=i+6The sum of the marks(the total) is 162t+i=162

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

t+i=162(i+6)+i=1622i=1626i=1562=78

This means that Ibironke's mark is 78.


STEP: Find Tshegofatso's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Ibironke's mark and one of the equations we have to find Tshegofatso's mark. The easiest way to do that is to substitute Ibironke's mark back into the first equation:

t=i+6=(78)+6=84

The students achieved these marks: Tshegofatso earned 84 and Ibironke earned 78.


Submit your answer as: and

Word problems: test scores and simultaneous equations

Chukwuma and Babalwe are friends. Chukwuma takes Babalwe's economics test paper and says: “I have 9 marks more than you do and the sum of both our marks is equal to 137. What are our marks?”

Answer:

Chukwuma got marks and Babalwe got marks for the economics test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

c= Chukwuma's markb= Babalwe's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

Chukwuma has 9 marks more than Babalwec=b+9The sum of the marks(the total) is 137c+b=137

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

c+b=137(b+9)+b=1372b=1379b=1282=64

This means that Babalwe's mark is 64.


STEP: Find Chukwuma's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Babalwe's mark and one of the equations we have to find Chukwuma's mark. The easiest way to do that is to substitute Babalwe's mark back into the first equation:

c=b+9=(64)+9=73

The students achieved these marks: Chukwuma earned 73 and Babalwe earned 64.


Submit your answer as: and

Word problems: test scores and simultaneous equations

Chidiebere and Adefoluke are friends. Chidiebere takes Adefoluke's civil technology test paper and says: “I have 13 marks more than you do and the sum of both our marks is equal to 133. What are our marks?”

Answer:

Chidiebere got marks and Adefoluke got marks for the civil technology test paper.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

c= Chidiebere's marka= Adefoluke's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

Chidiebere has 13 marks more than Adefolukec=a+13The sum of the marks(the total) is 133c+a=133

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

c+a=133(a+13)+a=1332a=13313a=1202=60

This means that Adefoluke's mark is 60.


STEP: Find Chidiebere's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Adefoluke's mark and one of the equations we have to find Chidiebere's mark. The easiest way to do that is to substitute Adefoluke's mark back into the first equation:

c=a+13=(60)+13=73

The students achieved these marks: Chidiebere earned 73 and Adefoluke earned 60.


Submit your answer as: and

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 32.

Find the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 32. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 32". Remember that sum means addition. So:

n1+n2=32

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=32, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=322n1+2=322n1=30n1=15

The result is n1=15. Notice that this means that the other number, n2, must be 17, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 32, and 15+17=32.

The smaller number is 15.


Submit your answer as:

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The sum of the numbers is 34.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of 34. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 34". Remember that sum means addition. So:

n1+n2=34

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=34, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=342n1+2=342n1=32n1=16

The result is n1=16. Notice that this means that the other number, n2, must be 18, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 34, and 16+18=34.

The smaller number is 16.


Submit your answer as:

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The sum of the numbers is 30.

Determine the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of 30. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 30". Remember that sum means addition. So:

n1+n2=30

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=30, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=302n1+2=302n1=28n1=14

The result is n1=14. Notice that this means that the other number, n2, must be 16, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 30, and 14+16=30.

The smaller number is 14.


Submit your answer as:

Word problems: checking answers

Ibironke is 20 years older than her sister, Tshegofatso. In 8 years, Ibironke will be 3 times as old as Tshegofatso. How old is Tshegofatso now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Tshegofatso is years old.

numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Tshegofatso's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Tshegofatso's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Ibironke and Tshegofatso. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aI=Ibironke's ageaT=Tshegofatso's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Ibironke is 20 years older than her sister, Tshegofatso." We need to translate that into mathematics. The key word is older, which tells us to use addition to relate the ages.

aI=aT+20

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 8 years, Ibironke will be 3 times as old as Tshegofatso." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aI+8=Ibironke's age in 8 yearsaT+8=Tshegofatso's age in 8 years

These are the ages at which Ibironke will be 3 times as old as Tshegofatso. We can put all this together as follows:

Ibironke's agein 8 years=3×Tshegofatso's agein 8 yearsaI+8=3(aT+8)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aI and aT. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aI=aT+20. We can substitute this into the second equation and solve for aT.

aI+8=3(aT+8)(aT+20)+8=3(aT+8)aT+28=3aT+244=2aT2=aT

Terrific: we have the answer.

Tshegofatso is 2 years old.

The correct answer is: 2.


Submit your answer as:

Word problems: checking answers

Ifetayo is 18 years older than her sister, Jessica. In 7 years, Ifetayo will be 10 times as old as Jessica. How old is Jessica now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Jessica is years old.

one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Jessica's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Jessica's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Ifetayo and Jessica. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aI=Ifetayo's ageaJ=Jessica's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Ifetayo is 18 years older than her sister, Jessica." We need to translate that into mathematics. The key word is older, which tells us to use addition to relate the ages.

aI=aJ+18

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 7 years, Ifetayo will be 10 times as old as Jessica." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aI+7=Ifetayo's age in 7 yearsaJ+7=Jessica's age in 7 years

These are the ages at which Ifetayo will be 10 times as old as Jessica. We can put all this together as follows:

Ifetayo's agein 7 years=10×Jessica's agein 7 yearsaI+7=10(aJ+7)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aI and aJ. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aI=aJ+18. We can substitute this into the second equation and solve for aJ.

aI+7=10(aJ+7)(aJ+18)+7=10(aJ+7)aJ+25=10aJ+7045=9aJ5=aJ

Terrific: we have the answer.

But wait a minute: aJ represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

Word problems: checking answers

Ifetayo is 20 years younger than her sister, Halima. In 7 years, Halima will be 21 times as old as Ifetayo. How old is Ifetayo now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Ifetayo is years old.

one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Ifetayo's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Ifetayo's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Halima and Ifetayo. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aH=Halima's ageaI=Ifetayo's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Ifetayo is 20 years younger than her sister, Halima." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aI=aH20

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 7 years, Halima will be 21 times as old as Ifetayo." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aH+7=Halima's age in 7 yearsaI+7=Ifetayo's age in 7 years

These are the ages at which Halima will be 21 times as old as Ifetayo. We can put all this together as follows:

Halima's agein 7 years=21×Ifetayo's agein 7 yearsaH+7=21(aI+7)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aH and aI. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aI=aH20. We can substitute this into the second equation and solve for aH.

aH+7=21(aI+7)aH+7=21((aH20)+7)aH+7=21aH273280=20aH14=aH

Terrific: this means that Halima is 14 years old. But the question asked for us to find Ifetayo's age. We can find it using the first equation, which relates the two ages:

aI=aH20=(14)20=6

So we finally got the answer to the question. But wait a minute: aI represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 35.

What is the value of the larger number?

Answer: The larger number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 35. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

n1=the smaller numberthis is the numberwe need to findn2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 35". Remember that product means multiplication. So:

n1n2=35

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

Remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)n2=35n222n2=35n222n235=0(n27)(n2+5)=0
n2=7 and n2=5

This solution led to two answers for n2. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n2=7.

This means the other number, n1, must be 5, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 35, and 75=35.

The larger number is 7.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 15.

Determine the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 15. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 15". Remember that product means multiplication. So:

n1n2=15

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=15, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=15n21+2n1=15n21+2n115=0(n1+5)(n13)=0
n1=5andn1=3

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=3.

This means the other number, n2, must be 5, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 15, and 35=15.

The smaller number is 3.


Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The product of the numbers is 80.

Find the value of the smaller number.

Answer: The smaller number is .
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of 80. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

this is the numberwe need to findn1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 80". Remember that product means multiplication. So:

n1n2=80

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=80, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=80n21+2n1=80n21+2n180=0(n1+10)(n18)=0
n1=10andn1=8

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=8.

This means the other number, n2, must be 10, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 80, and 810=80.

The smaller number is 8.


Submit your answer as:

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 182

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 182.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 182. That means if we multiply the numbers we get 182.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 182. In other words, if we multiply the numbers, we get 182. As an equation this is:

n1n2=182

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=182n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=182(n1)2+n1182=0(n1+14)(n113)=0
n1=14andn1=13

The solutions to the equation are n1=14 or n1=13. This means the first number is either −14 or 13.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −14 and 13 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=13, so the first number is n1=14.

Based on that we can find the second number. The second number is n1+1, which is equal to (14)+1=13.

The two consecutive integers are −14 and −13.


Submit your answer as: and

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are positive
  • the product of the numbers is 20

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 20.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 20. That means if we multiply the numbers we get 20.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 20. In other words, if we multiply the numbers, we get 20. As an equation this is:

n1n2=20

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=20n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=20(n1)2+n120=0(n1+5)(n14)=0
n1=5andn1=4

The solutions to the equation are n1=5 or n1=4. This means the first number is either −5 or 4.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −5 and 4 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the n1=5, so the first number is n1=4.

Based on that we can find the second number. The second number is n1+1, which is equal to (4)+1=5.

The two consecutive integers are 4 and 5.


Submit your answer as: and

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 182

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 182.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 182. That means if we multiply the numbers we get 182.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 182. In other words, if we multiply the numbers, we get 182. As an equation this is:

n1n2=182

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=182n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=182(n1)2+n1182=0(n1+14)(n113)=0
n1=14andn1=13

The solutions to the equation are n1=14 or n1=13. This means the first number is either −14 or 13.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −14 and 13 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=13, so the first number is n1=14.

Based on that we can find the second number. The second number is n1+1, which is equal to (14)+1=13.

The two consecutive integers are −14 and −13.


Submit your answer as: and

Word problems: an age-old question

David has a son, Fezekile. Here are some facts about how old David and Fezekile are:

  • David is 5 times as old as Fezekile right now.
  • 8 years from now, David will be 3 times as old as Fezekile.

How old are David and Fezekile now?

Answer:

David is years old and Fezekile is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: David and his son, Fezekile. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let d=David's ageLet f=Fezekile's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "David is now 5 times as old as Fezekile." As an equation, this is:

Ages now: d=5f

The second piece of information says that in "8 years... David will be 3 times as old as his son." In 8 years David will be d+8 years old, and similarly Fezekile will be f+8 years old. Then:

Ages in 8 years: d+8=3(f+8)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel d.)

d+8=3(f+8)(5f)+8=3f+242f=16f=8

Great! Now we know that Fezekile is 8 years old.


STEP: Use Fezekile's age to find David's age
[−1 point ⇒ 0 / 6 points left]

We can now find David's age. Substitute Fezekile's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

d=5f=5(8)=40

Write your final answer: David is 40 years old and Fezekile is 8 years old.


Submit your answer as: and

Word problems: an age-old question

Chibueze has a son, Adam. Here are some facts about how old Chibueze and Adam are:

  • Chibueze is 9 times as old as Adam right now.
  • 4 years from now, Chibueze will be 5 times as old as Adam.

How old are Chibueze and Adam now?

Answer:

Chibueze is years old and Adam is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Chibueze and his son, Adam. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let c=Chibueze's ageLet a=Adam's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Chibueze is now 9 times as old as Adam." As an equation, this is:

Ages now: c=9a

The second piece of information says that in "4 years... Chibueze will be 5 times as old as his son." In 4 years Chibueze will be c+4 years old, and similarly Adam will be a+4 years old. Then:

Ages in 4 years: c+4=5(a+4)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel c.)

c+4=5(a+4)(9a)+4=5a+204a=16a=4

Great! Now we know that Adam is 4 years old.


STEP: Use Adam's age to find Chibueze's age
[−1 point ⇒ 0 / 6 points left]

We can now find Chibueze's age. Substitute Adam's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

c=9a=9(4)=36

Write your final answer: Chibueze is 36 years old and Adam is 4 years old.


Submit your answer as: and

Word problems: an age-old question

Bandile has a son, Johan. Here are some facts about how old Bandile and Johan are:

  • Bandile is 7 times as old as Johan right now.
  • 4 years from now, Bandile will be 5 times as old as Johan.

How old are Bandile and Johan now?

Answer:

Bandile is years old and Johan is years old.

numeric
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Bandile and his son, Johan. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let b=Bandile's ageLet j=Johan's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Bandile is now 7 times as old as Johan." As an equation, this is:

Ages now: b=7j

The second piece of information says that in "4 years... Bandile will be 5 times as old as his son." In 4 years Bandile will be b+4 years old, and similarly Johan will be j+4 years old. Then:

Ages in 4 years: b+4=5(j+4)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel b.)

b+4=5(j+4)(7j)+4=5j+202j=16j=8

Great! Now we know that Johan is 8 years old.


STEP: Use Johan's age to find Bandile's age
[−1 point ⇒ 0 / 6 points left]

We can now find Bandile's age. Substitute Johan's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

b=7j=7(8)=56

Write your final answer: Bandile is 56 years old and Johan is 8 years old.


Submit your answer as: and

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "One Metre at a Time" marathon in Kimberley. The second-place runner finished 18,8 seconds after the winner. The third runner was 12,8 seconds behind the second runner, and the fourth runner was 23,99 seconds behind the third runner. If the second-place runner had an average speed of 19,19 km/h, what was the winning time for the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. Negative time is not possible: it must be a positive number.

The correct answer is: No, it cannot be negative.


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Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "One Metre at a Time" marathon in Port Elizabeth. The second-place runner finished 17,29 seconds after the winner. The third runner was 19,8 seconds behind the second runner, and the fourth runner was 28,17 seconds behind the third runner. If the second-place runner had an average speed of 19,73 km/h, how many of the runners took more than 2,17 hours to finish the marathon?

Can the answer to the question above be a non-integer number?

Answer:

Can the answer be non-integer?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be non-integer
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: how many of the runners took more than 2,17 hours to finish the marathon. So the question is about the number of people who finished the race after 2,17 hours had passed. The answer cannot be a non-integer. The number of people to finish after 2,17 hours might be 3 or 7 people, but it cannot be 4,26 people. In other words, the number of people must be a whole number.

The correct answer is: No, it must be an integer.


Submit your answer as:

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42,195 kilometres. 1 000 people ran the "One Metre at a Time" marathon in Kimberley. The second-place runner finished 16,6 seconds after the winner. The third runner was 17,48 seconds behind the second runner, and the fourth runner was 28,73 seconds behind the third runner. If the second-place runner had an average speed of 19,23 km/h, what was the winning time for the marathon?

Can the answer to the question above be a non-integer number?

Answer:

Can the answer be non-integer?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be non-integer
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. And non-integer values are acceptable for time values. (In other words, an amount of time does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.


Submit your answer as:

Word problems: rectangle facts

The diagonal of a rectangle is 26 cm more than its width. The length of the same rectangle is 13 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

the diagonal of the rectangle is26 cm more than its widthd=w+26the length of the rectangle is13 cm more than its widthl=w+13

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+26 in for d and w+13 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+26)2=w2+(w+13)2w2+52w+676=w2+(w2+26w+169)0=w226w507

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w226w5070=(w39)(w+13)w=39or w=13

This means that the width of the rectangle is either 39 cm or 13 cm. But the dimensions of a rectangle cannot be negative, so w=39 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+13=(39)+13=52 cm

The width of the rectangle is 39 cm and the length is 52 cm.


Submit your answer as: and

Word problems: rectangle facts

The diagonal of a rectangle is 28 cm more than its width. The length of the same rectangle is 14 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

the diagonal of the rectangle is28 cm more than its widthd=w+28the length of the rectangle is14 cm more than its widthl=w+14

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+28 in for d and w+14 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+28)2=w2+(w+14)2w2+56w+784=w2+(w2+28w+196)0=w228w588

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w228w5880=(w42)(w+14)w=42or w=14

This means that the width of the rectangle is either 42 cm or 14 cm. But the dimensions of a rectangle cannot be negative, so w=42 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+14=(42)+14=56 cm

The width of the rectangle is 42 cm and the length is 56 cm.


Submit your answer as: and

Word problems: rectangle facts

The diagonal of a rectangle is 25 cm more than its width. The length of the same rectangle is 17 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

the diagonal of the rectangle is25 cm more than its widthd=w+25the length of the rectangle is17 cm more than its widthl=w+17

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+25 in for d and w+17 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+25)2=w2+(w+17)2w2+50w+625=w2+(w2+34w+289)0=w216w336

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w216w3360=(w28)(w+12)w=28or w=12

This means that the width of the rectangle is either 28 cm or 12 cm. But the dimensions of a rectangle cannot be negative, so w=28 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+17=(28)+17=45 cm

The width of the rectangle is 28 cm and the length is 45 cm.


Submit your answer as: and

Setting up simultaneous equations

Last week, Anathi and Bokamoso had a chemistry test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 145.
  • Anathi's mark is 31 less than Bokamoso's mark.

Let a represent Anathi's mark and b represent Bokamoso's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 145.
Anathi's mark is 31 less than Bokamoso's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Anathi's mark and b for Bokamoso's mark. So we can break up the first fact like this:

The sum of the marksis145a+b=145

The first fact is equivalent to this equation: a+b=145.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Anathi's markis31 less than Bokamoso's marka=b31

This equation, a=b31, means that Anathi's mark is less that Bokamoso's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 145. a+b=145
Anathi's mark is 31 less than Bokamoso's mark. a=b31

Submit your answer as: and

Setting up simultaneous equations

Last week, Anathi and Bukelwa had a physics test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 159.
  • Anathi's mark is 19 more than Bukelwa's mark.

Let a represent Anathi's mark and b represent Bukelwa's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 159.
Anathi's mark is 19 more than Bukelwa's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Anathi's mark and b for Bukelwa's mark. So we can break up the first fact like this:

The sum of the marksis159a+b=159

The first fact is equivalent to this equation: a+b=159.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Anathi's markis19 more than Bukelwa's marka=b+19

This equation, a=b+19, means that Anathi's mark is more that Bukelwa's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 159. a+b=159
Anathi's mark is 19 more than Bukelwa's mark. a=b+19

Submit your answer as: and

Setting up simultaneous equations

Last week, Azubuike and Babangida had a maths test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 158.
  • Azubuike's mark is 6 more than Babangida's mark.

Let a represent Azubuike's mark and b represent Babangida's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 158.
Azubuike's mark is 6 more than Babangida's mark.
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Azubuike's mark and b for Babangida's mark. So we can break up the first fact like this:

The sum of the marksis158a+b=158

The first fact is equivalent to this equation: a+b=158.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Azubuike's markis6 more than Babangida's marka=b+6

This equation, a=b+6, means that Azubuike's mark is more that Babangida's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 158. a+b=158
Azubuike's mark is 6 more than Babangida's mark. a=b+6

Submit your answer as: and